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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 2 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 2
If $$x$$ is very large, then $$\dfrac {2x}{1+x}$$ is
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close to $$0$$
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arbitrarily large
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lie between $$2$$ and $$3$$
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close to $$2$$
Explanation
$$\dfrac {2x}{1+x}=\dfrac {2}{1+\dfrac {1}{x}}=\dfrac{2}{1+0}=2$$
If $$x\rightarrow \alpha$$ then $$\dfrac {1}{x}\rightarrow 0$$.
What is $$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \cos { x } }{ \pi -x } } $$ equal to?
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$$0$$
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$$\pi $$
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$$\dfrac { 1 }{ \pi } $$
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$$1$$
Explanation
Here, putting $$x=0$$ directly we have,
$$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \cos { x } }{ \pi -x } } $$
$$=\dfrac {\cos 0 }{ \pi -0 } =\dfrac { 1 }{ \pi } $$
Hence, C is correct.
$$\underset{x \rightarrow 0}{lim}(1+ax)^{\large{\frac{b}{x}}}$$ is equal to
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$$e^{ab}$$
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$$e^{a+b}$$
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non-existent
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none of these
Explanation
Now,
$$\underset{x \rightarrow 0}{lim}(1+ax)^{\large{\frac{b}{x}}}$$
$$=\underset{x \rightarrow 0}{lim}\left\{(1+ax)^{\large{\frac{1}{ax}}}\right\}^{ab}$$
$$=\left\{\underset{x \rightarrow 0}{lim}(1+ax)^{\large{\frac{1}{ax}}}\right\}^{ab}$$
$$=e^{ab}$$. [ Since $$\lim\limits_{x \to 0}(1+x)^{\large{\frac{1}{x}}}=e$$].
$$\displaystyle{\lim_{x \to 0}}$$ $$\Bigg(\dfrac{(1+x)^{2}}{e^{x}}\Bigg)^\dfrac{4}{\sin x}$$ is:
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$$e^2$$
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$$e^{4}$$
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$$e^8$$
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$$e^{-8}$$
Explanation
Now
$$\lim\limits_{x \to 0}\left(\dfrac{(1+x)^2}{e^{x}}\right)^{\dfrac{4}{\sin x}}$$
$$=\lim\limits_{x \to 0}\dfrac{\left(\left\{(1+x)^{\dfrac{1}{x}}\right\}^{\dfrac{x}{\sin x}}\right)^8}{e^{\cfrac{4x}{\sin x}}}$$
We have $$\lim\limits_{x \to 0}\dfrac{\sin x}{x}=1$$ and
$$\lim\limits_{x \to 0}(1+x)^{\dfrac{1}{x}}=e$$.
As both the limits of the numerator and denominator exists,and the limit of the numerator is non-vanishing, so we can write the above limits as
$$=\dfrac{\lim\limits_{x\to 0}\left(\left\{(1+x)^{\dfrac{1}{x}}\right\}^{\lim\limits_{x\to 0}\dfrac{x}{\sin x}}\right)^8}{\lim\limits_{x\to 0}e^{\cfrac{4x}{\sin x}}}$$ [Using division property of limits]
$$=\dfrac{\left(\lim\limits_{x\to 0}\left\{(1+x)^{\dfrac{1}{x}}\right\}^{\left(\lim\limits_{x\to 0}\dfrac{x}{\sin x}\right)}\right)^8}{e^{\left(\lim\limits_{x\to 0}\dfrac{4x}{\sin x}\right)}}$$ [Using limit property]
$$=\dfrac{(e^1)^8}{e^{\frac{4}{1}}}=e^2$$
If a sequence $$< a_{n} >$$ is such that $$a_{1},a_{n+1}=\dfrac {2+3a_{n}}{1+2a_{n}}$$ and $$\displaystyle \lim_{n \rightarrow \infty}a_{n}$$ exists, then $$a_{n}$$ is equal to
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$$\cos 36^{o}$$
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$$2\cos 36^{o}$$
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$$\sin 18^{o}$$
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$$2\sin 36^{o}$$
$$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$$ equals
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$$\dfrac{7}{3}$$
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$$\dfrac{10}{3}$$
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$$\dfrac{14}{3}$$
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$$\dfrac{1}{3}$$
Explanation
We have,
$$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$$
$$\lim_{x\rightarrow\ 0}\dfrac{\sin7x}{\sin3x}$$
$$\lim_{x\rightarrow\ 0}\dfrac{\dfrac{\sin7x}{7x}\times 7x}{\dfrac{\sin3x}{3x}\times 3x}$$
$$\lim_{x\rightarrow\ 0}\dfrac{\dfrac{\sin7x}{7x}\times 7}{\dfrac{\sin3x}{3x}\times 3}$$
$$=\dfrac{1\times 7}{1\times 3}$$
$$=\dfrac{7}{3}$$
Hence, this is the answer.
$$\displaystyle \lim_{x\rightarrow 0}{(1+\sin x)^{\cos x}}$$ is equal to
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$$0$$
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$$e$$
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$$1$$
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$$\dfrac{1}{e}$$
Explanation
$$\displaystyle{\lim}_{x\rightarrow 0}{\left(1+\sin{x}\right)}^{\cos{x}}$$
$$={\left(1+\sin{0}\right)}^{\cos{0}}$$
$$={1}^{1}=1$$
'
Evaluate the following limit :
$$lim_{x\rightarrow 0} \dfrac{1-\cos 2x}{x^2}$$
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$$0$$
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$$1$$
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$$2$$
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none of these
Explanation
$$lim_{x\rightarrow 0} \dfrac{1-\cos 2x}{x^2}$$
$$=lim_{x\rightarrow 0} \dfrac{2\sin^2 x}{x^2}$$
$$=2lim_{x \rightarrow 0} (\dfrac{\sin x}{x} \times \dfrac{\sin x}{x})$$
$$=2lim_{x\rightarrow 0} \dfrac{\sin x}{x} \times lim_{x\rightarrow 0} \dfrac{\sin x}{x}=2(1)(1)=2$$
Evaluate
$$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$$
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$$\cfrac { 1 }{ 2 } $$
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$$1$$
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$${ 3 }^{ 2 }$$
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$$\cfrac { 1 }{ { 2 }^{ 2 } } $$
Explanation
$$\displaystyle \lim_{n\rightarrow \infty} \dfrac{1+2+3+...+n}{n^2}$$
$$\displaystyle =\lim_{n\rightarrow \infty} \dfrac{1}{n^2} \times \dfrac{n(n+1)}{2}$$
$$\displaystyle =\lim_{n\rightarrow \infty} \dfrac{1}{2}(1+\dfrac{1}{n})=\dfrac{1}{2}$$
Which of the following statement is not correct
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$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) +g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } +\lim _{ x\rightarrow c }{ g\left( x \right) } $$
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$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) -g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } -\lim _{ x\rightarrow c }{ g\left( x \right) } $$
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$$\displaystyle \lim _{ x\rightarrow c }{ \left[ f\left( x \right) .g\left( x \right) \right] } =\lim _{ x\rightarrow c }{ f\left( x \right) } .\lim _{ x\rightarrow c }{ g\left( x \right) } $$
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$$\displaystyle \lim _{ x\rightarrow c }{ \dfrac { f\left( x \right) }{ g\left( x \right) } } =\displaystyle \dfrac { \lim _{ x\rightarrow c }{ f\left( x \right) } }{ \lim _{ x\rightarrow c }{ g\left( x \right) } } $$
Explanation
All the result other than D are correct.
The last result will hold only when $$g(x)\ne 0$$ and $$\lim\limits_{x\to c} g(x)\ne 0$$ for $$x$$ belonging to a neighbourhood of $$c$$.
A, B, C are general results which are always true.
What is the value of $$\underset{x\rightarrow 0}{lim}\dfrac{\sin\,x}{\tan \,3x}$$
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$$\dfrac{1}{4}$$
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$$\dfrac{1}{3}$$
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$$\dfrac{1}{2}$$
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$$1$$
Explanation
Given,
$$\lim_{x\rightarrow 0}\dfrac{\sin \left(x\right)}{\tan \left(3x\right)}$$
applying L'Hospital rule, we get,
$$\lim_{x\rightarrow 0} \dfrac{\cos \left(x\right)}{\sec ^2\left(3x\right)\cdot 3}$$
$$=\dfrac{\cos \left(0\right)}{\sec ^2\left(3\cdot 0\right)\cdot 3}$$
$$=\dfrac{1}{3}$$
If $$f(x)=\left\{\begin{array}{l}x-5;\>x\leq 1\\4x^{2}-9;\>1<x\le2\\3x^{2}+4;\>x>2\end{array}\right.$$
Then $$f(2^{+})-f(2^{-})=$$
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$$0$$
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$$2$$
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$$9$$
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$$4$$
Explanation
Given: $$f(x)=\left\{\begin{array}{l}x-5;\>x\leq 1\\4x^{2}-9;\>1<x\le2\\3x^{2}+4;\>x>2\end{array}\right.$$
$$f(2^{+})=\displaystyle \lim _{x\rightarrow 2^+} f(x)$$
$$=\displaystyle \lim _{ h\rightarrow 0}3(2+h)^{2}+4 $$
$$f(2^+)=16$$
$$f(2^{-})=\displaystyle \lim _{x\rightarrow 2^-} f(x)$$
$$=\displaystyle \lim _{ h\rightarrow 0}4(2-h)^{2}-9$$
$$f(2^-)=7$$
So, $$f(2^{+})-f(2^{-})=16-7=9$$
If $$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}2\mathrm{x}+\mathrm{b}(\mathrm{x}<\alpha)\\\mathrm{x}+\mathrm{d}(\mathrm{x}\geq\alpha)\end{array}\right.$$is such that
$$\displaystyle \lim_{x\rightarrow \alpha}f(x) =L$$, then $$L=$$
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$$2d-b$$
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$$b-d$$
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$$d+b$$
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$$b-2d$$
Explanation
$$\begin{matrix}lim\\h\rightarrow 0^{-} \end{matrix}\ f(x)=2(\alpha -h)+b=2\alpha +b=L .............(1)\ $$
$$\begin{matrix}lim\\h\rightarrow 0 ^{+} \end{matrix}\ f(x)=(\alpha +h)+d=L\quad \quad \ \alpha =L-d\ ................(2)$$
Substituting value of euation (2)in (1), we get
$$2\left ( L-d \right )+b=L$$
$$L=2d-b$$
Hence, option 'A' is correct.
$$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \frac{\pi x}{2})=$$
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$$\pi$$
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$$ 2\pi$$
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$$\displaystyle \frac{\pi}{2}$$
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$$\displaystyle \frac{2}{\pi}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \dfrac{\pi x}{2})$$
$$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { 1-x }{ \cot { (\dfrac { \pi x }{ 2 } } ) } $$
It is of the form $$\displaystyle \dfrac{0}{0}$$, so applying L-Hospital's rule
$$=\lim _{ x\rightarrow 1 } \displaystyle\dfrac { -1 }{ -\dfrac { \pi }{ 2 } \csc ^{ 2 }{ (\dfrac { \pi x }{ 2 } ) } } $$
$$=\displaystyle \dfrac{2}{\pi}$$
lf $$f(x)=x,x<0;f(x)=0,x=0$$; $$f(x)=x^{2};x>0$$, then $$\displaystyle \lim_{x\rightarrow 0}f(x)$$ is equal to
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Does not exist
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$$0$$
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$$-1$$
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$$1$$
Explanation
Given: $$f(x)$$= $$\left\{ \begin{matrix} x \\ 0 \\ x^ 2 \end{matrix}\begin{matrix}\quad x<0 \\\quad x=0 \\ \quad x>0 \end{matrix} \right\} $$
To Find : $$\underset{x\rightarrow 0}{\lim}$$ $$f(x)$$= ?
Sol: left hand limit $$\rightarrow $$
$$\underset{x\rightarrow 0^-}{\lim}$$ $$f(x)$$=
$$\underset{x\rightarrow 0}{\lim}$$ $$x$$ =$$0$$
right hand limit
$$\rightarrow $$
$$\underset{x\rightarrow 0^+}{\lim}$$
f(x)=
$$\underset{x\rightarrow 0}{\lim}$$ $$x^2$$ =$$0$$,
LHL=RHL
$$f(0)=0$$
$$LHL=RHL=f(0)=0$$
$$f(x)=2x+1, a=1,l=3$$ and $$\epsilon=0.001$$, then $$\delta>0$$ satisfying $$0<|x-a|<\delta$$ such that $$|f(x)-l|<\epsilon$$, is
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$$0.0005$$
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$$0.005$$
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$$0.001$$
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$$0.0001$$
Explanation
We have to find $$\delta>0$$, which satisfies $$0<|x-a|<\delta$$ such that $$|f(x)-l|<\epsilon$$
$$\because |f(x)-l| <\epsilon$$
$$\Rightarrow -\epsilon <f(x)-l<\epsilon$$
$$\Rightarrow l-\epsilon<f(x)<l+\epsilon$$
$$\Rightarrow 3-\epsilon<2x+1<3+\epsilon$$
$$\Rightarrow -\epsilon<2x-2<\epsilon$$
$$\Rightarrow -\dfrac{\epsilon }{2}<x-1<\dfrac{\epsilon }{2}$$
$$\Rightarrow |x-1|<\dfrac{\epsilon }{2}$$
$$\Rightarrow |x-a| <\dfrac{\epsilon }{2}$$
$$\delta = \dfrac{\epsilon}{2} = \dfrac{0.001}{2} = 0.0005$$
Hence, option A.
Evaluate: $$\displaystyle \lim_{h\rightarrow 0}\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right)$$
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$$\displaystyle \frac{1}{12}$$
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$$-\displaystyle \frac{4}{3}$$
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$$-\displaystyle \frac{16}{3}$$
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$$-\displaystyle \frac{1}{48}$$
Explanation
Let $$L =\displaystyle \lim_{h\rightarrow 0}\left(\frac{1}{h\sqrt[3]{8+h}}-\frac{1}{2h}\right) =\lim_{h\rightarrow 0}\frac{1}{h}\left(\frac{1}{\sqrt[3]{8+h}}-\frac{1}{2}\right)$$
Let $$a= \sqrt[3]{8+h}\Rightarrow a^3 = 8+h$$ and $$b=2\Rightarrow b^3 =8$$
Also $$a^3-b^3=(a-b)(a^2+b^2+ab)=8+h-8 =h\Rightarrow $$ (1)
$$\therefore\displaystyle L = \lim_{h\to 0}\frac{1}{h}\left( \frac{1}{a}-\frac{1}{b}\right)=-\lim_{h\to 0}\frac{a-b}{h}\cdot \frac{1}{ab}$$
$$\displaystyle \quad = -\lim_{h\to 0}\frac{a^3-b^3}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}$$
$$\displaystyle \quad = -\lim_{h\to 0}\frac{h}{h}\cdot \frac{1}{ab(a^2+b^2+ab)}$$ .....using $$(1)$$
As $${h \rightarrow 0}\Rightarrow a^3=8\Rightarrow a=2$$
$$\displaystyle \quad =-\frac{1}{2\times 2(2^2+2^2+2\times 2)}=$$
$$=-\dfrac{1}{48}$$
lf $$\displaystyle \lim_{x\rightarrow a^+}f(x)=L$$, then for each $$\epsilon>0$$, there exists $$\delta>0$$ so that
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$$ 0<|x-a|<\delta\Rightarrow |f(x)-L|\geq\epsilon$$
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$$ 0<|x-a|<\delta\Rightarrow |f(x)-L|<\epsilon$$
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$$a < x < a+\delta\Rightarrow f(x) - L <\epsilon$$
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$$ a-\delta < x < a\Rightarrow |f(x)-L| < \epsilon$$
Explanation
It is fundamental concept that, for limit of a function $$f(x)$$ to exist at any point $$a$$ there exists a real number $$\delta>0$$, such that $$0< |x-a|<\delta$$, for which $$|f(x)-L| < \epsilon$$, where $$\epsilon >0$$
$$\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos x}{x\log(1+x)}=$$
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$$1$$
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$$0$$
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$$-1$$
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$$\displaystyle \dfrac{1}{2}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\dfrac{1-\cos x}{x\log(1+x)}$$
$$\displaystyle\lim _{ x\rightarrow 0 } \displaystyle \dfrac { 2\sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ x\log (1+x) }$$
$$ =\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { 2\displaystyle\dfrac { \sin ^{ 2 }{ \dfrac { x }{ 2 } } }{ { \left(\displaystyle\dfrac { x }{ 2 } \right) }^{ 2 } } \times \dfrac { 1 }{ 4 } }{\displaystyle \dfrac { 1 }{ x } \log (1+x) } $$
$$=\displaystyle\lim _{ x\rightarrow 0 }\displaystyle \dfrac { \displaystyle\dfrac { 1 }{ 2 } }{ \dfrac{\log{ (1+x) }}{ x } } =\dfrac{1}{2}$$
$$\displaystyle \lim_{x\rightarrow 0}\frac{(1-e^{x})\sin x}{x^{2}+x^{3}}=$$
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$$-1$$
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$$0$$
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$$1$$
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$$2$$
Explanation
The expression can be simplified as,
(lim $$ x \rightarrow 0 \dfrac{(1-{e}^{x})}{x}) \times (\dfrac{sin x}{x})( lim _{ x \rightarrow 0} \dfrac{1}{1+x} )$$
Now using,
lim $$ x \rightarrow 0 \dfrac{({e}^{x} - 1)}{x} = 1 $$
lim $$ x \rightarrow 0 \dfrac{sin x}{x} $$ $$= 1$$
We get,
$$(-1)$$ $$ \times 1 \times lim x \rightarrow 0 \dfrac{1}{1+x} $$
$$= -1$$
lf $$f(x)=2x-3,a=2,l=1$$ and $$\epsilon =0.001$$ then $$\delta>0$$ satisfying$$ 0<|x-a|<\delta, \ \ |f(x)-l|<\epsilon$$, is:
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$$0.005$$
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$$0.0005$$
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$$0.001$$
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$$0.0001$$
Explanation
$$|f(x)-l|<0.001=\epsilon$$
$$\Rightarrow |2x-3-1| < 0.001$$
$$\Rightarrow -0.001<2x-4<0.001$$
$$\Rightarrow -0.0005<x-2<0.0005$$
$$\Rightarrow |x-2|<0.0005$$
$$\Rightarrow |x-a|<0.0005=\delta$$
Hence, $$\delta = 0.0005 > 0$$
$$\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=$$
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$$-\dfrac{1}{2}$$
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$$\dfrac{1}{2}$$
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$$1$$
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$$\dfrac{3}{2}$$
Explanation
We simplify the given expression as,
$$\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}({e}^{x - sin x} - 1)}{2(x - sin x)} $$
Let $$x - sin x = y$$
As $$ x \rightarrow 0 \text{ so does y } \rightarrow 0 $$
Hence, the question transforms into
$$(\displaystyle\lim_{x \rightarrow 0} \dfrac{{e}^{sin x}}{2}) \times (\lim _{y \rightarrow 0}\dfrac{{e}^{y} - 1}{y}) $$
$$= \dfrac{1}{2} \times 1 $$
$$= \dfrac{1}{2} $$
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}$$=
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$$2$$
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$$-2$$
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$$\displaystyle \frac{1}{2}$$
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$$-\displaystyle \frac{1}{2}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}$$
$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { x\dfrac { 2\tan { x } }{ 1-\tan ^{ 2 }{ x } } -2x\tan x }{\left(1-\dfrac { 1-\tan ^{ 2 }{ x } }{ 1+\tan ^{ 2 }{ x } } \right)^{ 2 } } } $$
$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x\tan ^{ 3 }{ x } }{ 4\tan ^{ 4 }{ x } }}$$
$$\displaystyle =\lim _{ x\rightarrow 0 }{ \dfrac { 2x }{ 4\tan { x } } } $$
$$=\displaystyle \dfrac { 1 }{ 2 } $$
lf $$ \displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) $$ is finite then the value of $$a,b$$ respectively are
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$$5\, -4$$
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$$-5,\ -4$$
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$$-4,\ 3$$
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$$4,\ 5$$
Explanation
$$\lim _{ x\rightarrow 0 } \left(\displaystyle \frac { \cos 4x+a\cos 2x+b }{ x^{ 4 } } \right) $$
As $${ x\rightarrow 0 }$$, denominator tends to 0, so the numerator also tends to 0.
$$\Rightarrow \lim _{ x\rightarrow 0 } \cos 4x+a\cos 2x+b=0$$
at $$x = 0 \Rightarrow \cos 4x = 1, \cos 2x = 1$$
$$\Rightarrow 1+a+b=0$$
$$\Rightarrow a+b=-1$$
Now, put $$a = -4 $$
$$ \Rightarrow b = 3$$
Option C satisfies above equation.
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}$$=
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{3}{2}$$
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$$\displaystyle \frac{3}{4}$$
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$$\displaystyle \frac{1}{4}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}=\lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos x)(1+\cos x+\cos^2x)}{x\sin 2x}$$
$$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{2\sin^2\frac{x}{2}(1+\cos x+\cos^2x)}{4x\cos x\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\lim_{x\rightarrow 0}\displaystyle \dfrac{\sin\dfrac{x}{2}(1+\cos x+\cos^2x)}{4 \times \dfrac x2\cos \tfrac x2 \times \cos x } =\frac{3}{4}$$...... As$$\left \{ \lim_{\tfrac x2 \to0} \dfrac {\sin \tfrac x2}{\tfrac x2}=0 \right \}$$
$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}$$=
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1
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0
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-1
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$$\displaystyle \dfrac{1}{\pi}$$
Explanation
$$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { (\dfrac { \pi }{ 2 } -x)\sec x }{ \csc { x } } $$
$$\displaystyle=\lim _{ h\rightarrow 0 } \dfrac { (\dfrac { \pi }{ 2 } -\dfrac { \pi }{ 2 } +h)\sec { (\dfrac { \pi }{ 2 } -h) } }{ \csc { (\dfrac { \pi }{ 2 } -h } ) }$$
$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\csc { h } }{ \sec { h } } $$
$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { h\cos { h } }{ \sin { h } } =1$$
Solve :
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$$
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$$\displaystyle \frac{3}{4}$$
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$$\displaystyle \frac{1}{4}$$
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$$\displaystyle \frac{4}{3}$$
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$$0$$
Explanation
Solve : $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$$
$$=\underset{x\rightarrow0}\lim\dfrac{\sin x}{x}\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}+x)\times\underset{x\rightarrow 0}\lim\sin(\dfrac{\pi}{3}-x)$$
$$=1\times\sin\dfrac{\pi}{3}\times\sin\dfrac{\pi}{3}$$
$$=1\times\dfrac{\sqrt3}{2}\times\dfrac{\sqrt3}{2}$$
$$=\dfrac34$$
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}=$$
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$$\dfrac{10}{3}$$
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$$\dfrac{3}{10}$$
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$$\dfrac{6}{5}$$
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$$\dfrac{5}{6}$$
Explanation
Using, $$1 - cos 2x = 2{sin}^{2} x $$,
The expression transforms to,
$$lim _{ x \rightarrow 0 } \dfrac{ 2{sin}^{2} xsin 5x}{{x}^{2}sin 3x} $$
Rewriting the expression in a different form,
$$lim _{ x \rightarrow 0 } \dfrac{2{sin}^{2} x}{{x}^{2}} \times \dfrac{sin 5x}{5x} \times \dfrac{3x}{sin 3x} \times \dfrac{5}{3} $$
Therefore, the limit to the expression is $$ \dfrac{10}{3} $$
Hence, option 'A' is correct.
The value of $$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } $$ is:
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$$0$$
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$$1$$
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$$\displaystyle \frac { \sin { \beta } }{ \beta } $$
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$$\displaystyle \frac { \sin { 2\beta } }{ 2\beta } $$
Explanation
$$\displaystyle \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \beta } }{ { \alpha }^{ 2 }-{ \beta }^{ 2 } } \right] } =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } \sin { \left( \alpha +\beta \right) } }{ \left( \alpha -\beta \right) \left( \alpha +\beta \right) } \right] } $$
$$\displaystyle =\lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } \times \lim _{ \alpha \rightarrow \beta }{ \left[ \frac { \sin { \left( \alpha +\beta \right) } }{ \left( \alpha +\beta \right) } \right] } $$
$$\displaystyle =\lim _{ \alpha -\beta \rightarrow 0 }{ \left[ \frac { \sin { \left( \alpha -\beta \right) } }{ \left( \alpha -\beta \right) } \right] } .\left( \frac { \sin { 2\beta } }{ 2\beta } \right) $$ ....... $$[\because \alpha \rightarrow \beta \implies \alpha - \beta \rightarrow 0]$$
$$\displaystyle =1.\frac { \sin { 2\beta } }{ 2\beta }$$ ..... $$[\because \displaystyle \lim_{x\rightarrow 0} \dfrac{\sin x}{x}=1]$$
$$=\dfrac { \sin { 2\beta } }{ 2\beta } $$
Evaluate: $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$$
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$$\displaystyle \frac{3}{2}$$
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$$\displaystyle \frac{2}{3}$$
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$$\displaystyle \frac{1}{3}$$
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$$\displaystyle \frac{3}{4}$$
Explanation
Given,
$$\lim_{x\rightarrow 0}\dfrac{\sec 4x-\sec 2x}{\sec 3x-\sec x}$$
$$\sec x=\dfrac{1}{\cos x}$$
$$=\displaystyle\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{\cos 4x}-\dfrac{1}{\cos 2x}}{\dfrac{1}{\cos 3x}-\dfrac{1}{\cos x}}$$
$$=\lim_{x\rightarrow 0}\dfrac{\cos 2x-\cos 4x}{\cos x-\cos 3x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$
$$=\lim_{x\rightarrow 0}\dfrac{-2\sin 3x \sin (-x)}{-2\sin (2x)\sin (-x)}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$
$$=\lim_{x\rightarrow 0}\dfrac{\dfrac{\sin 3x}{3x}\times 3x}{\dfrac{\sin 2x}{2x}\times 2x}\dfrac{\cos 3x\cos x}{\cos 4x\cos 2x}$$
as $$\cos 0=1$$ and $$\lim_{x\rightarrow 0}\dfrac{\sin x}{x}=1$$
$$=\dfrac{3x}{2x}=\dfrac{3}{2}$$
$$\displaystyle \lim_{x \rightarrow\frac{\pi}{2}}\displaystyle \dfrac{cosecx-\cot x}{x}$$=
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$$1$$
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$$\displaystyle \frac{2}{\pi}$$
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$$\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{1}{5}$$
Explanation
$$\displaystyle \lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { \csc { x } -\cot x }{ x } $$
$$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 2 } } \dfrac { 1-\cos x }{ x\sin { x } } $$
$$\displaystyle =\dfrac{2}{\pi}$$
$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \frac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$$=
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$$\sqrt{2}$$
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$$\displaystyle \frac{1}{\sqrt{2}}$$
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$$-\sqrt{2}$$
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$$\displaystyle \frac{-1}{\sqrt{2}}$$
Explanation
$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{4}}\displaystyle \dfrac{\sec x.\tan(4x-\pi)}{\sin(4x-\pi)}$$
$$\displaystyle =\lim _{ x\rightarrow \dfrac { \pi }{ 4 } } \dfrac { \sec x.\dfrac { \tan (4x-\pi ) }{ (4x-\pi ) } }{ \dfrac { \sin (4x-\pi ) }{ (4x-\pi ) } }$$
$$ =\sqrt { 2 } $$
$$\displaystyle \lim_{x\rightarrow \frac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=$$
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$$\sqrt{3}$$
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$$\dfrac{1}{\sqrt{3}}$$
0%
$$-\sqrt{3}$$
0%
$$-\dfrac{1}{\sqrt{3}}$$
Explanation
$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{6}}\frac{3\sin x-\sqrt{3}\cos x}{6x-\pi }=2\sqrt{3}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\dfrac{\sqrt{3}}{2}\sin x-\dfrac{1}{2}\cos x}{6x-\pi }$$
$$=\displaystyle \dfrac{2\sqrt{3}}{6}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\cos\dfrac{\pi}{6}\sin x-\sin\dfrac{\pi}{6}\cos x}{x-\dfrac{\pi}{6} }$$
$$=\displaystyle \dfrac{1}{\sqrt{3}}\lim_{x\rightarrow \dfrac{\pi }{6}}\dfrac{\sin \left(x-\dfrac{\pi}{6}\right)}{\left(x-\dfrac{\pi}{6} \right)}=\dfrac{1}{\sqrt{3}}$$
$$\displaystyle \lim_{x\rightarrow 0}\frac{tan x^{0}}{x}=$$
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0
0%
1
0%
$$\displaystyle \frac{\pi}{180}$$
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$$\pi$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\dfrac{tan x^{0}}{x}$$
$$\displaystyle={ \dfrac { \pi }{ 180 } }\lim _{ x\rightarrow 0 } \dfrac { \tan { (\dfrac { \pi }{ 180 } x) } }{ { (\dfrac { \pi }{ 180 } ) }x } $$
$$\displaystyle={ \dfrac { \pi }{ 180 } }$$
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$$=
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$$0$$
0%
$$1$$
0%
$$(\displaystyle \frac{\pi}{180})^{3}$$
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$$4.(\displaystyle \frac{\pi}{180})^{3}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\sin x-\sin 3x}{x^{3}}$$
$$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ x } }{ x^{ 3 } }$$
$$\displaystyle =\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 } } $$
$$\displaystyle ={ (\frac { \pi }{ 180 } ) }^{ 3 }\lim _{ x\rightarrow 0 } \frac { 4\sin ^{ 3 }{ (\frac { \pi }{ 180 } x) } }{ x^{ 3 }{ (\frac { \pi }{ 180 } ) }^{ 3 } } $$
$$\displaystyle =4{ (\frac { \pi }{ 180 } ) }^{ 3 }$$
Solve:
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{3\tan x-\tan 3x}{2x^{3}}$$
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$$\dfrac{1}{4}$$
0%
$$\dfrac{3}{4}$$
0%
$$4$$
0%
$$-4$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3\tan x-\tan 3x}{2x^{3}}$$
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{3(x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+......)-(3x+\dfrac{(3x)^3}{3}+\dfrac{2(3x)^5}{15}+....)}{2x^{3}}=\dfrac{1-9}{2}=-4$$
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}$$=
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$$\displaystyle \log(\frac{a}{b})$$
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$$\displaystyle \log(\frac{b}{a})$$
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$$\log(ab)$$
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$${\log_{b}}{a}$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}$$
$$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-1-b^{x}+1}{{x}}$$
$$=\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(a^{x}-1)-(b^{x}-1)}{{x}}$$
$$=\log a - \log b$$ ......... Using formula
$$=\log\left(\dfrac{a}{b}\right)$$
Hence,
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{a^{x}-b^{x}}{{x}}=\log\left(\dfrac{a}{b}\right)$$
$$\displaystyle \lim_{x\rightarrow 0}(\frac{\sin x-x}{x})(\sin\frac{1}{x})$$ is:
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does not exist
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is equal to 0
0%
is equal to 1
0%
exists and different from 0 and 1
Explanation
$$sin x = x -\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}....(1)$$
$$sin\ x-x =-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!}$$
$$\dfrac{sin\ x-x}{x} =-\dfrac{x^{2}}{3!}+\dfrac{x^{4}}{5!}-\dfrac{x^{6}}{7!}....=0\ (as\ x\ found\ to\ 0)$$
sin$$(\dfrac{1}{x})$$ is an oscillatory function means same finite no.
$$0\times finite\ no.=0$$
$$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$$=
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$$2$$
0%
$$1$$
0%
$$0$$
0%
$$3$$
Explanation
$$\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\cos x-\sin x}{(\frac{\pi}{4}-x)(\cos x+\sin x)}$$
$$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{1-\tan x}{(\frac{\pi}{4}-x)(1+\tan x)}$$.................(dividing the num and deno by cosx)
$$=\displaystyle \lim_{x\rightarrow \frac{\pi }{4}}\displaystyle \frac{\tan\left(\frac{\pi}{4}-x\right) }{(\frac{\pi}{4}-x)} = \lim_{h\to 0}\frac{\tan h}{h} = 1$$
$$\displaystyle \lim_{x\rightarrow \displaystyle \frac{\pi }{2}}\displaystyle \frac{1-\sin\theta}{\cos\theta\left(\dfrac{\pi}{2}-{\theta}\right)}=$$
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$$1$$
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$$-1$$
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$$-\displaystyle \frac{1}{2}$$
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$$\displaystyle \frac{1}{2}$$
Explanation
$$Put\ \theta = \dfrac{\pi }{2}-\phi $$
$$\begin{matrix}Lim\\\phi \rightarrow 0 \end{matrix}\ \dfrac{1-cos\phi}{\dfrac{sin\phi}{\phi}\phi ^{2}}=\dfrac{1}{2}$$
$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$$=
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$$\displaystyle \frac{1}{3}$$
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$$\displaystyle \frac{1}{4}$$
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$$\displaystyle \frac{1}{6}$$
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$$\displaystyle \frac{1}{8}$$
Explanation
$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \frac{1-\sin x}{(\pi-2x)^{2}}$$
$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\sin (\dfrac { \pi }{ 2 } -h) }{ (\pi -2(\dfrac { \pi }{ 2 } -h))^{ 2 } } $$......(x-->h-pi/2)
$$\displaystyle =\lim _{ h\rightarrow 0 } \dfrac { 1-\cos { h } }{ 4h^{ 2 } }$$
$$\displaystyle =\dfrac { 1 }{ 8 } $$
$$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}=$$
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$$1$$
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$$-1$$
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$$\dfrac{1}{3}$$
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$$0$$
Explanation
$$\displaystyle \lim_{x\rightarrow \infty }\frac{x+\sin x}{x+ \cos x}$$
$$=\displaystyle \lim_{x\rightarrow \infty }\frac{\displaystyle 1+\frac{\sin x}{x}}{\displaystyle 1+ \frac{\cos x}{x}}$$
$$=\displaystyle \frac{\displaystyle 1+\frac{\text{Any finite number between -1 and 1}}{\infty}}{\displaystyle 1+ \frac{\text{Any finite number between -1 and 1}}{\infty}}=\frac{1+0}{1+0}=1$$
$$ \displaystyle f(x)=\left\{ \begin{matrix} \dfrac { 3 }{ { x }^{ 2 } } \sin { 2{ x }^{ 2 } } ,\; x<0 \\ \dfrac { { x }^{ 2 }+2x+C }{ 1-3{ x }^{ 2 } } ,\; x\ge 0,\; x\neq \frac { 1 }{ \sqrt { 3 } } \\ 0,\; x=\frac { 1 }{ \sqrt { 3 } } \end{matrix} \right. $$
$$\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=$$
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$$\frac{C}{4}$$
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2C
0%
$$\frac{C}{3}$$
0%
C
Let $$f(x)=\begin{cases}x^{2}-1,0 < x < 2\\2x+3,2 \leq x < 3\end{cases},$$
The quadratic equation whose roots are $$\displaystyle \lim_{x\rightarrow 2^{-}}f(x)$$ and $$\displaystyle \lim_{x\rightarrow 2^{+}}f(x)$$ is
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$$x^{2}-10x+21=0$$
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$$x^{2}-6x+9=0$$
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$$x^{2}-14x+49=0$$
0%
$$x^{2}+6x+9=0$$
Explanation
For $$0 < x < 2$$,
$$f(x) = {x}^{2} - 1 $$
$$\lim_{ x \rightarrow {2}^{-}} f(x) = 3 $$
For, $$2 < x < 3$$
$$f(x) = 2x + 3$$
$$\lim_{ x \rightarrow {2}^{+}} f(x) = 7 $$
Thus the equation with roots, 3 and 7 is
$$ {x}^{2} - 10x + 21 = 0 $$
The value of $$\sqrt{e}$$
e
upto four decimal places is?
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$$1.6484$$
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$$1.5237$$
0%
$$1.2589$$
0%
$$1.9190$$
Explanation
$$(e)^{\frac {1}{2}}=1+\dfrac {1}{2}+\dfrac {(\dfrac {1}{2})^2}{2!}+\dfrac {(\dfrac {1}{2})^3}{3!}+\dfrac {(\dfrac {1}{2})^4}{4!}$$
$$=1+0\cdot 5+0\cdot 125+0\cdot 02083+0\cdot 00260$$
$$=1\cdot 6484$$
If $$f(x) = \displaystyle \frac {x^2+6x}{\sin x}$$ , then $$\displaystyle \lim_{x\rightarrow 0^{-}}f(x)=$$
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$$2$$
0%
$$4$$
0%
$$6$$
0%
$$8$$
Explanation
$$\lim_{x\rightarrow 0 }^{ - }f\left( x \right) =\dfrac { { x }^{ 2 }+6x }{ \sin x } \\= \lim_{ h\rightarrow 0 } f\left( h \right) $$
$$=\dfrac { { h }^{ 2 }+6(0-h) }{ \sin(0-h) }= \lim_{h\rightarrow 0 }f\left( h \right) =\dfrac { { h }-6 }{ -\sin(h)/h } = \dfrac { -6 }{ -1 } =6$$
Hence, the answer is "C".
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\log(1+ax)-\log(1+bx)}{x}$$=
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$$a+b$$
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$$a-b$$
0%
$$-(a+b)$$
0%
$$ab$$
Explanation
$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{\log(1+ax)-\log(1+bx)}{x}$$
$$=\displaystyle a.\lim_{x\to 0}\frac{\log(1+ax)}{ax}-b.\lim_{x\to 0}\frac{\log(1+bx)}{bx}=a-b$$
Let $$f$$ be a continuous function on [1,3]. lf $$f$$ takes only rational values for all $$x$$ and $$f(2)=10$$ then $$f(1.5)$$ is equal to
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$$8$$
0%
$$\displaystyle \frac{f(1)+f(3)}{3}$$
0%
$$20$$
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$$10$$
Explanation
Since $$f$$ is continuous, so it must take all real values between $$f\left(1\right)$$ and $$f\left(3\right)$$.
But since $$f$$ takes only rational values so $$f$$ must be a constant function.
Hence $$f\left(1.5\right)=f\left(2\right)=10$$
Evaluate: $$\displaystyle \underset{x\rightarrow 0}{\lim}\ \ \frac{\sin3x^{2}}{\cos(2x^{2}-x)}$$
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$$0$$
0%
$$-1$$
0%
$$4$$
0%
$$-6$$
Explanation
Given,
$$\lim _{x\to \:0}\left(\dfrac{\sin \left(3\right)x^2}{\cos \left(2x^2-x\right)}\right)$$
$$=\dfrac{\sin \left(3\right)\cdot \:0^2}{\cos \left(2\cdot \:0^2-0\right)}$$
$$=0$$
$$\displaystyle \lim _{ x\rightarrow { 0 }^{ - } }{ \dfrac { 3\sin(2{ x }^{ 2 }) }{ { x }^{ 2 } } } =A$$
then the value of $$A$$ is
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$$2$$
0%
$$4$$
0%
$$6$$
0%
$$8$$
Explanation
$$\displaystyle \lim_{x\rightarrow { 0^- }} \dfrac { 3\sin({ 2x }^{ 2 }) }{ { x }^{ 2 } } =\lim_{x\rightarrow { 0^- }}\dfrac { 2*3\sin({ 2x }^{ 2 }) }{ { 2x }^{ 2 } } = 6$$
Hence, answer is option $$C$$
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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