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CBSE Questions for Class 11 Engineering Maths Limits And Derivatives Quiz 9 - MCQExams.com
CBSE
Class 11 Engineering Maths
Limits And Derivatives
Quiz 9
If
π
∫
0
x
f
(
s
i
n
x
)
d
x
=
A
π
/
2
∫
0
f
(
s
i
n
x
)
d
x
,
then A is _____________.
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0%
0
0%
π
0%
π
/
4
0%
2
π
lim
x
→
∞
[
n
n
2
+
1
2
+
n
n
2
+
2
2
+
n
n
2
+
3
2
+
.
.
.
.
+
1
n
5
]
Report Question
0%
π
/
4
0%
tan
−
1
(
2
)
0%
π
/
2
0%
tan
−
1
(
3
)
Find the derivative of
√
tan
x
with respect to
x
using the first principle.
Report Question
0%
s
e
c
2
x
2
√
t
a
n
x
0%
s
e
c
x
2
√
t
a
n
x
0%
s
e
c
2
x
√
t
a
n
x
0%
s
e
c
2
x
2
√
t
a
n
Explanation
y
=
√
tan
x
⇒
d
y
d
x
=
1
2
√
tan
x
×
sec
2
x
=
sec
2
x
2
√
tan
x
Hence, the answer is
sec
2
x
2
√
tan
x
.
The value of
n
l
i
m
→
∞
1.
n
+
2.
(
n
−
1
)
+
3.
(
n
−
2
)
+
.
.
.
+
n
.1
1
2
+
2
2
+
.
.
.
+
n
2
is
Report Question
0%
1
0%
−
1
0%
1
√
2
0%
1
2
Explanation
Now,
1.
n
+
2
(
n
−
1
)
+
3.
(
n
−
2
)
+
.
.
.
.
+
n
.1
The general term of the above series is
T
r
=
r
(
n
−
r
+
1
)
=
r
(
n
+
1
−
r
)
Sum of series
=
S
=
n
∑
r
=
1
T
r
=
n
∑
r
=
1
r
(
n
+
1
−
r
)
S
=
n
∑
r
=
1
n
r
+
n
∑
r
=
1
r
−
n
∑
r
=
1
r
2
=
n
n
∑
r
=
1
r
+
n
∑
r
=
1
r
−
n
∑
r
=
1
r
2
=
n
n
(
n
+
1
)
2
+
n
(
n
+
1
)
2
−
n
(
n
+
1
)
(
2
n
+
1
)
6
Similarly,
1
2
+
2
2
+
3
2
+
.
.
.
.
.
+
n
2
=
n
(
n
+
1
)
(
2
n
+
1
)
6
lim
n
→
∞
1.
n
+
2
(
n
−
1
)
+
3.
(
n
−
2
)
+
.
.
.
+
n
.1
1
2
+
2
2
+
3
2
+
.
.
.
.
.
.
.
.
n
2
lim
n
→
∞
n
2
(
n
+
1
)
2
+
n
(
n
+
1
)
2
−
n
(
n
+
1
)
(
2
n
+
1
)
6
n
(
n
+
1
)
(
2
n
+
1
)
6
lim
n
→
∞
n
2
+
1
2
−
2
n
+
1
6
2
n
+
1
6
lim
n
→
∞
(
3
n
+
3
−
2
n
−
1
2
n
+
1
)
=
lim
n
→
∞
(
3
+
3
n
−
2
−
1
n
2
+
1
n
)
=
3
−
2
2
=
1
2
The value of
lim
x
→
0
1
+
sin
x
−
cos
x
+
log
(
1
−
x
)
x
3
, is
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0%
−
1
0%
1
/
2
0%
−
1
/
2
0%
1
The value of
lim
θ
→
0
+
sin
√
θ
√
sin
θ
is equal to
Report Question
0%
0
0%
1
0%
−
1
0%
4
f
(
x
)
=
sin
x
and
f
′
(
π
)
Report Question
0%
−
1
0%
0
0%
1
0%
None of these
Explanation
f
(
x
)
=
sin
x
f
′
(
x
)
=
d
d
x
sin
x
f
′
(
x
)
=
cos
x
f
′
(
π
)
=
cos
π
=
−
1
If
x
+
y
=
sin
(
x
−
y
)
then
d
y
d
x
is equal to
Report Question
0%
1
2
0%
0
0%
−
1
0%
1
3
Explanation
Given,
x
+
y
=
sin
(
x
−
y
)
x
+
y
=
sin
x
cos
y
−
cos
x
sin
y
1
+
d
y
d
x
=
sin
x
(
−
sin
y
)
+
cos
y
(
cos
x
)
−
cos
x
(
cos
y
)
−
(
−
sin
x
)
sin
y
1
+
d
y
d
x
=
−
sin
x
sin
y
+
cos
x
cos
y
+
sin
x
sin
y
−
cos
x
cos
y
1
+
d
y
d
x
=
0
∴
If
y = \sqrt{\dfrac{sec x-1}{sec x+1}}
then
\dfrac{dy}{dx}
=
Report Question
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\dfrac{1}{2} sec^2 \dfrac{x}{2}
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sec^2 \dfrac{x}{2}
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\dfrac{1}{2} tan \dfrac{x}{2}
0%
tan \dfrac{x}{2}
Explanation
Given,
y=\sqrt{\dfrac{\sec x-1}{\sec x+1}}
=\sqrt{\dfrac{\sec x-1}{\sec x+1}} \times \sqrt{\dfrac{\sec x-1}{\sec x-1}}
=\sqrt{\dfrac {(sec x-1)^2}{(\sec x-1)(\sec x+1)}}
=\sqrt{\dfrac{(sec x-1)^2}{\sec ^2x-1}}
=\sqrt{\dfrac{(sec x-1)^2}{\tan ^2x}}
=\dfrac{sec x-1}{\tan x}
=\dfrac{sec x}{\tan x}-\dfrac{1}{\tan x}
=\csc x -\cot x
\therefore y=\csc x -\cot x
Now,
\dfrac{dy}{dx}=\dfrac{d}{dx}(\csc x -\cot x)
=-\cot \left(x\right)\csc \left(x\right)-\left(-\csc ^2\left(x\right)\right)
=-\cot \left(x\right)\csc \left(x\right)+\csc ^2\left(x\right)
=\left(\dfrac{1}{\sin \left(x\right)}\right)^2-\dfrac{1}{\sin \left(x\right)}\cdot \dfrac{\cos \left(x\right)}{\sin \left(x\right)}
=\dfrac{1}{\sin ^2\left(x\right)}-\dfrac{\cos \left(x\right)}{\sin ^2\left(x\right)}
=\dfrac{1-\cos \left(x\right)}{\sin ^2\left(x\right)}
=\dfrac{1-\cos \left(x\right)}{1-\cos ^2\left(x\right)}
=-\dfrac{1-\cos \left(x\right)}{\left(\cos \left(x\right)+1\right)\left(\cos \left(x\right)-1\right)}
=\dfrac{1}{\cos \left(x\right)+1}
=\dfrac{1}{2 \cos^2\left(\dfrac{x}{2} \right)-1+1}
=\dfrac{1}{2} sec^2 \dfrac{x}{2}
Let
f:(0, \infty)\to R
be a differentiable function such that
f'(x)=2-\dfrac{f(x)}{x}
for all
x\in (0, \infty)
and
f(1)\neq 1
. Then
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\underset { x\rightarrow { 0 }^{ + } }{ \lim } f'\left( \dfrac { 1 }{ x } \right) =1
0%
\underset { x\rightarrow { 0 }^{ + } }{ \lim } xf\left( \dfrac { 1 }{ x } \right) =2
0%
\underset { x\rightarrow { 0 }^{ + } }{ \lim } x^{ 2 }f'\left( x \right) =0
0%
\left| f\left( x \right) \right| \le 2
for
all
X\in \left( 0,2 \right)
Explanation
Here,
f'(x)=2-\dfrac{f(x)}{x}
or
\dfrac{dy}{dx}+\dfrac{y}{x}=2
[ i.e. linear differential equation in
y
]
Integrating Factor,
IF=e^{\displaystyle \int{\dfrac{1}{x}dx}}=e^{\log x}=x
\therefore
Required solution is
y.(IF)=\displaystyle \int{Q(IF)dx+C}
\Rightarrow y(x)=\int{2(x)dx+C}
\Rightarrow yx=x^2+C
\therefore y=x+\dfrac{C}{x}
[ \because C \neq 0
, as
f(1) \neq 1]
(a)
\displaystyle \lim_{x \rightarrow 0^+}{f' \left( \dfrac{1}{x}\right)}=\displaystyle \lim_{x \rightarrow 0^+}{(1-Cx^2)}=1
\therefore
Option (a) is correct.
(b)
\displaystyle \lim_{x \rightarrow 0^+}{xf \left( \dfrac{1}{x}\right)}=\displaystyle \lim_{x \rightarrow 0^+}{(1+Cx^2)}=1
\therefore
Option (b) is correct.
(c)
\displaystyle \lim_{x \rightarrow 0^+}{x^2f' (x)}=\displaystyle \lim_{x \rightarrow 0^+}{(x^2-C)}=-C\neq 0
\therefore
Option (c) is correct.
(d)
f(x)=x+\dfrac{C}{x}, C \neq 0
For
C > 0, \displaystyle \lim_{x\rightarrow 0^+}{f(x)=\infty}
\therefore
Function is not bounded in
(0, 2)
.
\therefore
Option (d) is incorrect.
If
\mathrm { L } = \lim _ { \mathrm { x } ^ { 2 } \rightarrow \mathrm { a } } \frac { \mathrm { b } - \cos \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) } { \left( \mathrm { x } ^ { 2 } - \mathrm { a } \right) \sin \left( \mathrm { cx } ^ { 2 } - \mathrm { a } \right) }
is non-
zero finite
( \mathrm { a } > 0 ) ,
then-
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0%
L = 2 , b = 1 , c = 1
0%
L = \frac { 1 } { 2 } , b = 1 , c = 1
0%
L = 4 , b = - 1 , c = - 1
0%
L = \frac { 1 } { 4 } , b = - 1 , c = - 1
The solution the differential equation
\cos x \sin y dx+ \sin x \cos y dy =0
Report Question
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\dfrac{\sin x}{\sin y}=c
0%
\cos x+ \cos y=c
0%
\sin x + \sin y =c
0%
\sin x. \sin y=c
Explanation
Given differential equation is
\cos x \sin ydx+\sin x \cos ydy=0
\Rightarrow \cos x\sin y dx=-\sin x\cos y dy
\Rightarrow \dfrac{\cos x}{\sin x}dx=-\dfrac{\cos y}{\sin y}dy
On integrating both sides, we get
\int {\cot xdx}=\int{-\cot ydy}
\log \sin x=-\log \sin y+\log C
\Rightarrow \log \sin x \sin y=\log C
\Rightarrow \sin x. \sin y=C
For
x>y
,
\displaystyle\lim_{x\rightarrow 0}{\left[\left(\sin{x}\right)^{1/x}+\left(\cfrac{1}{x}\right)^{\sin{x}}\right]}
is :
Report Question
0%
0
0%
-1
0%
1
0%
2
If the function
f(x)
satisfies the relation
f(x+y)=y\dfrac{|x-1|}{(x-1)}f(x)+f(y)
with
f(1)=2
, then
\displaystyle\lim_{x\rightarrow 1}f'(x)
is?
Report Question
0%
2
0%
-2
0%
0
0%
Limit do not exixst
Let
f : R \to R
be a differentiable function satisfying
f'(3) + f'(2) = 0
.
Then
\underset{x \to 0}{\lim} \left(\dfrac{1+f(3+x)-f(3)}{1+f(2-x) - f(2)}\right)^{\frac{1}{x}}
is equal to
Report Question
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e^2
0%
e
0%
e^{-1}
0%
1
Explanation
\underset{x \to 0}{\lim} \left(\dfrac{1+f(3+x)-f(3)}{1+f(2-x) - f(2)}\right)^{\frac{1}{x}}
(1^{\infty}
form
)
\Rightarrow e^{\underset{x \to 0}{\lim} \dfrac{f(3+x)-f(2-x)-f(3)+f(2)}{x(1+f(2-x)-f(2))}}
using L'Hopital
\Rightarrow e^{\underset{x \to 0}{\lim}\dfrac{f'(3+x)+f'(2-x)}{-xf'(2-x)+(1+f(2-x)-f(2))}}
\Rightarrow e^{\dfrac{f'(3) + f'(2)}{1}} = 1
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow a}\dfrac{\sqrt{x}+\sqrt{a}}{x+a}
.
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-\dfrac{1}{\sqrt{a}}
0%
\dfrac{1}{{a}}
0%
\dfrac{1}{2\sqrt{a}}
0%
\dfrac{1}{\sqrt{a}}
Explanation
Evaluate
\displaystyle \lim_{x\to a}\dfrac {\sqrt x +\sqrt a}{x+a}
not an indeterminant Form so, limit can be found by simply putting unit of
x
\displaystyle \lim_{x\to a}\dfrac {\sqrt x +\sqrt a}{x+a}=\dfrac {2\sqrt a}{2a}=\dfrac {1}{\sqrt a}
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 0}\dfrac{3x+1}{x+3}
.
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\dfrac{1}{3}
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\dfrac{2}{3}
0%
\dfrac{5}{3}
0%
None of these
Explanation
Evaluate
\displaystyle \lim _{ x\rightarrow 0} \dfrac {3x+1}{x+3}
as
\to 0, Expression \to \dfrac {1}{3}
, not indeterminant form
so, Limit can be foord, by simple substituting value
\displaystyle \lim _{ x\rightarrow 0} \dfrac {3x+1}{x+3}=\dfrac {1}{3}
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 0}\dfrac{x^{2/3}-9}{x-27}
.
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\dfrac{1}{3}
0%
\dfrac{1}{2}
0%
\dfrac{1}{5}
0%
None of these
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 0}\dfrac{ax+b}{cx+d}, d\neq 0
.
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\dfrac{a}{c}
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\dfrac{a}{d}
0%
\dfrac{b}{d}
0%
None of these
Explanation
\displaystyle \lim_{x \to 0}\dfrac {ax+b}{cx+d}, d\neq 0
as
x\to 0
, Expression
\to \dfrac {b}{d}
, which is a determined Form, so Limit= value
\displaystyle \lim_{x \to 0}\dfrac {ax+b}{cx+d}=\dfrac {b}{a}
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 1}\dfrac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}, x > 1
.
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\dfrac{\sqrt{2}+1}{\sqrt{2}}
0%
\dfrac{\sqrt{2}-1}{\sqrt{2}}
0%
\dfrac{\sqrt{2}+1}{{2}}
0%
None of these
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a^2+x^2}-a}{x^2}
.
Report Question
0%
\dfrac{1}{\sqrt a}
0%
\dfrac{1}{\sqrt {2a}}
0%
\dfrac{1}{a}
0%
\dfrac{1}{2a}
Explanation
\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a^{2}+x^{2}}-a}{x^{2}}
As
x\rightarrow 0
, it is
\dfrac{0}{0}
Form,
So, using rationalisation
\displaystyle\lim_{x\rightarrow 0}\dfrac{(\sqrt{a^{2}+x^{2}}-a)}{x^{2}}\times \dfrac{(\sqrt{a^{2}+x^{2}}+a)}{\sqrt{a^{2}+x^{2}}+a)}
\displaystyle\lim_{x\rightarrow 0}\dfrac{(a^{2}+x^{2})-a^{2}}{x^{2}(\sqrt{a^{2}+x^{2}}+a)}
((a+b)(a+b)=a^{2}-b^{2})
\displaystyle\lim_{x\rightarrow 0}\dfrac{x^{2}}{x^{2}(\sqrt{a^{2}+x^{2}}+a)}
\displaystyle\lim_{x\rightarrow 0}\dfrac{1}{(\sqrt{a^{2}+x^{2}}+a)}=\dfrac {1}{\sqrt {a^2}+a}=\dfrac{1}{2a}
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 0}\dfrac{2x}{\sqrt{a+x}-\sqrt{a-x}}
.
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-2\sqrt{a}
0%
\sqrt{a}
0%
2\sqrt{a}
0%
None of these
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow a}\dfrac{x-a}{\sqrt{x}-\sqrt{a}}
.
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2\sqrt{a}
0%
2{a}
0%
2{a^{\frac 13}}
0%
None of these
Explanation
\displaystyle \lim_{x\rightarrow a}{\dfrac{x-a}{\sqrt x-\sqrt a}}
if
x\rightarrow a
, expression
\rightarrow \dfrac{0}{0}
, an indetermined form
using factorisation
\displaystyle \lim_{x\rightarrow a}{\dfrac{(\sqrt{x}-\sqrt a)(\sqrt x+\sqrt a)}{(\sqrt x-\sqrt a)}}=2\sqrt a
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a^2+ax}}
.
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\dfrac{1}{2\sqrt{a}}
0%
\dfrac{1}{2a\sqrt{a}}
0%
\dfrac{1}{2a}
0%
None of these
Explanation
\displaystyle \lim_{x \rightarrow 0} \dfrac{\sqrt{a+x}-\sqrt{a}}{x \sqrt{a^{2}+ ax}}
as
x \rightarrow 0
, it is
\dfrac{0}{0}
form
So, using rationalization
\displaystyle \lim_{x \rightarrow 0} \dfrac{\sqrt{a+x}- \sqrt{a}}{x \sqrt{a^{2}+ax}} \dfrac{(\sqrt{a+x}+ \sqrt{a})}{(\sqrt{a+x}+ \sqrt{a})}
\displaystyle \lim_{x \rightarrow 0} \dfrac{x}{x \sqrt{a^{2}+ax} (\sqrt{a+x}+ \sqrt{a})}
\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{ \sqrt{a^{2}+ax} (\sqrt{a+x}+ \sqrt{a})}
= \dfrac{1}{2a \sqrt{a}}
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 4}\dfrac{2-\sqrt{x}}{4-x}
.
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\dfrac{1}{4}
0%
\dfrac{1}{2}
0%
\dfrac{1}{3}
0%
None of these
Explanation
\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{4-\sqrt x}}
if
x\rightarrow 4
, expression
\rightarrow \dfrac{0}{0}
, an indetermined form
using factorisation
\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{2^2-(\sqrt x)^2}}=\displaystyle \lim_{x\rightarrow 4}{\dfrac{2-\sqrt x}{(2-\sqrt x)(2+\sqrt x)}}=\dfrac{1}{4}
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 2}\dfrac{\sqrt{1+4x}-\sqrt{5+2x}}{x-2}
.
Report Question
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\dfrac{1}{2}
0%
\dfrac{1}{3}
0%
\dfrac{1}{4}
0%
\dfrac{1}{5}
Explanation
Evaluate
\displaystyle \lim_{x\to 2}\dfrac {\sqrt {1+4x}-\sqrt {5+2x}}{(x-2)}
as
x\to 2
, it is
\dfrac {0}{0}
From so,
using rationalisation
\displaystyle \lim_{x\to 2}\dfrac {(\sqrt {1+4x}-\sqrt {5+2x}) (1+4x)+\sqrt {5+2x}}{(x-2) (\sqrt {1+4x}+\sqrt {5+2x})}
\displaystyle \lim_{x\to 2}\dfrac {((1+4x)-(5+2x))}{(x-2) (\sqrt {1+4x}+\sqrt {5+2x})}
\displaystyle \lim_{x\to 2}\dfrac {2(x-2)}{(x-2)(6)}=\dfrac {1}{3}
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 2}\dfrac{x-2}{\sqrt{x}-\sqrt{2}}
.
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0%
2\sqrt{3}
0%
2\sqrt{2}
0%
2\sqrt{5}
0%
None of these
Explanation
\displaystyle\lim_{x\rightarrow 2}\dfrac{(x-2)}{\sqrt{x}-\sqrt{2}}
as
x\rightarrow 2
, it is
\dfrac{0}{0}
Form,
So, using Factorisation
\displaystyle\lim_{x\rightarrow 2}\dfrac{((\sqrt x)^2-(\sqrt 2)^2)}{\sqrt{x}-\sqrt{2}}=\displaystyle\lim_{x\rightarrow 2}\dfrac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{(\sqrt{x}-\sqrt{2})}
\displaystyle \lim_{x\to 2}\sqrt x +\sqrt 2=2\sqrt{2}
Evaluate the following limit.
\displaystyle\lim_{x\rightarrow 0}\dfrac{8^x-2^x}{x}
.
Report Question
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log 4
0%
log 6
0%
log 5
0%
None of these
Explanation
we know
\displaystyle \lim_{x\rightarrow 0}{\dfrac{a^x-1}{x}}=\log a
\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-2^x}{x}}=\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-1+1-2^x}{x}}
=\displaystyle \lim_{x\rightarrow 0}{\dfrac{8^x-1}{x}}-\displaystyle \lim_{x\rightarrow 0}{\dfrac{2^x-1}{x}}=\log 8-\log 2
=\log 8/2=\log 4
.
Evaluate the following limits.
\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}
.
Report Question
0%
\dfrac{1}{\sqrt{2}}
0%
-\dfrac{1}{\sqrt{3}}
0%
-\dfrac{1}{{2}}
0%
-\dfrac{1}{\sqrt{2}}
Explanation
\displaystyle \lim_{x\rightarrow 0}{\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}}
As
x\rightarrow 0
, it is
\dfrac{0}{0}
from, so,
using rationalisation
\displaystyle \lim_{x\rightarrow 0}{\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}}\dfrac{(\sqrt{2-x}+\sqrt{2+x})}{(\sqrt{2-x}+\sqrt{2+x})}
\displaystyle \lim_{x\rightarrow 0}{\dfrac{(2-x)-(2+x)}{x(\sqrt{2-x}+\sqrt{2+x})}}
\displaystyle \lim_{x\rightarrow 0}{\dfrac{-2x}{x(\sqrt{2-x}+\sqrt{2+x})}}=\dfrac{-2}{2\sqrt{2}}=\dfrac{-1}{\sqrt2}
Evaluate the following limits.
If
\displaystyle\lim_{x\rightarrow a}\dfrac{x^9-a^9}{x-a}=9
, find all possible values of a.
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2, -2
.
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1, -1
.
0%
1, 0
.
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None of these
Explanation
Using formula
\displaystyle \lim_{x\rightarrow a}{\dfrac{x^n-a^n}{x-a}}=na^{n-1}
\displaystyle \lim_{x\rightarrow a}{\dfrac{x^9-a^9}{x-a}}=9.9^8=9
a^8=1\Rightarrow a=\pm 1
0:0:2
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