Limits And Continuity     - Class 11 Commerce Applied Mathematics - Extra Questions

The value of $$\displaystyle \lim_{x\rightarrow 0}x\,\sec x$$

$$L=\lim _{ x\rightarrow { 0 }^{ + } }{ \dfrac { |x| }{ x }  }$$

Let $$L=\lim _{ x\rightarrow { 0 }^{ + } }{ \dfrac { |x| }{ x }  }$$

$$\lim _{ x\rightarrow \infty  }{ { \left( \dfrac { 1+2x }{ 1+3x }  \right)  }^{ x } } \\ \\ =\lim _{ x\rightarrow \infty  }{ { \left( \dfrac { \dfrac { 1 }{ x } +2 }{ \dfrac { 1 }{ x } +3 }  \right)  }^{ x } } \\ =\lim _{ x\rightarrow \infty  }{ { \left( \dfrac { 2 }{ 3 }  \right)  }^{ x } } \\ =0$$

We have,

$$f\left( x \right)=\dfrac{\tan x}{x-\pi }$$

 

Since, $$\underset{x\to \pi }{\mathop{\lim }}\,f\left( x \right)$$

 

Now,

$$\underset{x\to \pi }{\mathop{\lim }}\,\left( \dfrac{\tan x}{x-\pi } \right)$$

 

This is the $$\dfrac{0}{0}$$ form.

 

So, apply L-Hospital rule,

$$\underset{x\to \pi }{\mathop{\lim }}\,\left( \dfrac{{{\sec }^{2}}x}{1-0} \right)$$

$$\underset{x\to \pi }{\mathop{\lim }}\,\left( {{\sec }^{2}}x \right)$$

$$\underset{x\to \pi }{\mathop{\lim }}\,{{\left( \sec x \right)}^{2}}$$

$$={{\left( \sec \pi  \right)}^{2}}$$

$$={{\left( -1 \right)}^{2}}$$

$$=1$$

 

Hence, this is the answer.

We have,

$$\displaystyle {{\left( \underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}} \right)}^{2}}$$  

We know that

$$\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\dfrac{1}{x} \right)}^{x}}=e$$

 Therefore,

$$={{e}^{2}}$$

Hence, this is the answer.

We have,

We have,

We have,

We have,

$$\underset{x\to a}{\mathop{\lim }}\,{{\left( \dfrac{x+1}{2x+1} \right)}^{{{x}^{2}}}}$$

Taking limit and we get,

$$\Rightarrow \underset{x\to a}{\mathop{\lim }}\,{{\left( \dfrac{x+1}{2x+1} \right)}^{{{x}^{2}}}}$$

$$\Rightarrow {{\left( \dfrac{a+1}{2a+1} \right)}^{{{a}^{2}}}}$$

Hence, this is the answer.

We have,

$$\lim _{ x\rightarrow \infty  }{ x\sin { (\cfrac { \pi  }{ 4x } ) } \cos { (\cfrac { \pi  }{ 4x } ) }  } \\ \lim _{ x\rightarrow \infty  }{ \cfrac { \sin { (\cfrac { \pi  }{ 4x } ) } \cos { (\cfrac { \pi  }{ 4x } ) }  }{ \cfrac { 1 }{ x }  }  }$$

$$\underset {x \rightarrow -2 } {\lim} \dfrac{tan \, \pi x}{x + 2}$$    ( $$0/0$$ form) 

Applying L'Hospital's Rule we get : 

$$\underset {x \rightarrow -2 } {\lim} \dfrac{\dfrac{d(tan\, \pi x)}{dx}}{\dfrac{d(x + 2)}{dx}}$$  

  $$\underset {x \rightarrow -2 } {\lim} \dfrac{\pi (sec^2 \pi x)}{1}$$  

Applying limit we get 
$$= \pi ( \because \, sec^2 (-2 \pi) = 1)$$

We have 
$$\underset {x \rightarrow 0} {\lim} \dfrac{e^x - 1}{x}$$  

$$\underset {x \rightarrow 0} {\lim} \left \{ \dfrac{ \left [ 1 + x + \dfrac{x^2}{2^1} + \dfrac{x^3}{3^1} + ----- \right ] - 1}{x} \right \}$$     $$\left [\because \, e^x = 1 + x + \dfrac{x^2}{2^1} + ----  \right ]$$  

$$\underset {x \rightarrow 0} {\lim} \left \{ \dfrac{x + \dfrac{x^2}{2^1} + \dfrac{x^3}{3^1} + -----}{x} \right \}$$  

$$\underset {x \rightarrow 0} {\lim} x \left \{ \dfrac{1 + \dfrac{x}{2^1} + \dfrac{x^2}{3^1} + -----}{x}  \right \}$$  

$$= 1 + 0 = 1$$

$$\underset {x \rightarrow 0 } {\lim}\dfrac{x \, tan \, 4x}{1 - cos \, 4x}$$  

$$= \underset {x \rightarrow 0 } {\lim}\dfrac{x \, sin \, 4x}{cos\,4x[2\,sin^22x]}$$  

$$=  \underset {x \rightarrow 0 } {\lim}\dfrac{2x\,sin\, 2s \, cos\, 2x}{cos \, 4x(2\,sin^22x)}$$  

$$= \underset {x \rightarrow 0 } {\lim} \left  ( \dfrac{cos\,2x}{cos\,4x} \dfrac{2x}{sin\,2x} \times \dfrac{1}{2}  \right  )$$  

$$= \dfrac{1}{2} \dfrac{\underset {2x \rightarrow 0} {\lim} cos\,2x}{\underset {4x \rightarrow 0} {\lim} cos\,4x} \times \underset {2x \rightarrow 0} {\lim} \left  ( \dfrac{2x}{sin\,2x}  \right  ) = \dfrac{1}{2} \times 1 = \dfrac{1}{2}$$

$$\underset {2h \rightarrow 0 } {\lim} \dfrac{e^{2h} - 1}{2h} \times 2$$  

$$= 1 \times 2 = 2$$

$$\displaystyle \underset{x\rightarrow \infty }{Lt}\dfrac{e^{x^{2}}-1}{e^{x^{2}}+1}$$
$$=\displaystyle \underset{x\rightarrow \infty }{Lt}\dfrac{1-\left ( \dfrac{1}{e} \right )^{x^{2}}}{1+\left ( \dfrac{1}{e} \right )^{x^{2}}}= \dfrac{1-0}{1+0}= 1.$$
$$\because \displaystyle e> 1 \therefore \dfrac{1}{e}< 1$$ and $$\displaystyle \left ( \dfrac{1}{e} \right )^{x^{2}}\rightarrow 0$$ as $$\displaystyle x\rightarrow \infty$$

Reqd. limit is equal to
$$=\displaystyle \lim_{x\rightarrow \pi /3}\frac{\tan x(\sqrt{3}-\tan x)(\tan x+\sqrt{3})}{\left (\frac{1}{2}\cos x\left ( \sqrt{3}-\tan x \right )  \right )}$$

$$=\displaystyle \lim_{x\rightarrow \pi /3}\frac{2\tan x(\tan x+\sqrt{3})}{\cos x}$$

$$=\displaystyle \frac{2(\sqrt{3})(\sqrt{3}+\sqrt{3})}{1/2}=24$$

$$\displaystyle \lim_{x\rightarrow \tan^{-1}3}\tan ^{4}x\frac{\tan ^{2}x-2\tan x-3}{\tan ^{2}x-4\tan x+3}$$

$$=\displaystyle \lim_{x\rightarrow \tan^{-1}3}\tan ^{4}x\frac{(\tan x-3)(\tan x+1)}{(\tan x-3)(\tan x-1)}$$

$$=\displaystyle \lim_{x\rightarrow \tan^{-1}3}\tan ^{4}x\frac{(\tan x+1)}{(\tan x-1)}=3^{4}\left(\frac{3+1}{3-1}\right)=162$$

Since f is differentiable, so by definition we have
$$\displaystyle f'\left( x \right) =\lim _{ h\rightarrow 0 }{ \frac { f\left( x+h \right) -f\left( x \right)  }{ h }  }$$
$$\displaystyle \Rightarrow f'\left( -x \right) =\lim _{ h\rightarrow 0 }{ \frac { f\left( -x+h \right) -f\left( -x \right)  }{ h }  } =\lim _{ h\rightarrow 0 }{ \frac { f\left( -\left( x-h \right)  \right) -f\left( -x \right)  }{ h }  }$$

$$\displaystyle \lim_{x\rightarrow 0}\frac{\sin \,ax}{\sin \,bx}=\lim_{x\rightarrow 0}\frac{\left ( \frac{\sin \,ax}{ax} \right )\times ax}{\left ( \frac{\sin \,bx}{bx} \right )\times bx}$$

$$\displaystyle \lim_{x\rightarrow 0}\frac{\cos \,2x-1}{\cos \,x -1}$$
$$=\displaystyle \lim_{x\rightarrow 0}\frac{\cos \,2x-1}{\cos \,x -1}=\lim_{x\rightarrow 0}\frac{1-2\sin ^{2}x-1}{1-2\sin ^{2}\frac{x}{2}-1},$$   

Using $$\displaystyle \left [ \cos \,x=1-2\sin ^{2}\frac{x}{2} \right ]$$
$$= \displaystyle \lim_{x\rightarrow 0}\frac{\sin ^{2}x}{\sin ^{2}\frac{x}{2}}=\lim_{x\rightarrow 0}\frac{\left ( \frac{\sin ^{2}x}{x^{2}} \right )\times x^{2}}{\left ( \frac{\sin ^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )\times \frac{x^{2}}{4}}$$
$$= 4\cdot \displaystyle \frac{\lim_{x\rightarrow 0}\left ( \frac{\sin ^{2}x}{x^{2}} \right )}{\lim_{x\rightarrow 0}\left ( \frac{\sin ^{2}\frac{x}{2}}{\left ( \frac{x}{2} \right )^{2}} \right )}=4$$

$$\displaystyle \lim_{x\rightarrow \frac{\pi }{2}}\frac{\tan 2x}{x-\frac{\pi }{2}}$$
Put  $$\displaystyle x-\frac{\pi }{2}=y$$ so that  $$\displaystyle x\rightarrow \frac{\pi }{2},\,y\rightarrow 0$$
$$\displaystyle \therefore \lim_{x\rightarrow \frac{\pi }{2}}\frac{\tan \,2x}{x-\frac{\pi }{2}}=\lim_{y\rightarrow 0}\frac{\tan 2\left ( y+\frac{\pi }{2} \right )}{y}$$
$$=\displaystyle \lim_{y\rightarrow 0}\frac{\tan \left ( \pi +2y \right )}{y}$$
$$=\displaystyle \lim_{y\rightarrow 0}\frac{\tan \,2y}{y}$$    
$$=\displaystyle \lim_{y\rightarrow 0}\frac{\tan \,2y}{2y}\cdot 2 =(1)\cdot 2=2$$

$$\displaystyle \lim_{x\rightarrow 0}\frac{ax+x\,\cos \,x}{b\,\sin \,x}=\frac{1}{b}\lim_{x\rightarrow 0}\frac{x\left ( a+\cos \,x \right )}{\sin \,x}$$
$$=\displaystyle \frac{1}{b}\lim_{x\rightarrow 0}\left ( \frac{x}{\sin \,x} \right )\times \lim_{x\rightarrow 0}\left ( a+\cos \,x \right )$$
$$\displaystyle = \frac{1}{b}\times \frac{1}{\left ( \lim_{x\rightarrow 0}\frac{\sin x}{x} \right )}\times \lim_{x\rightarrow 0}\left ( a+\cos x \right )$$
=  $$\displaystyle \frac{1}{b}\times \left ( a+\cos \,0 \right )= \frac{a+1}{b}$$

$$\lim\limits_{x\to0}\dfrac{\sin3x}{2x}$$

$$= \displaystyle \lim_{x\rightarrow 0} \dfrac {\cos \dfrac {x}{2} - \sin \dfrac {x}{2}}{\cos x}$$

$$lim_{x \rightarrow 2} \dfrac{x^5 - 32}{x^3 - 8}$$

$$\mathop {\lim }\limits_{x \to 0}$$ $$\dfrac { { \cos  x-\cot  x } }{ x }$$

$$= \mathop {\lim }\limits_{x \to 0}$$ $$\dfrac { { \cos  x-\dfrac { { \cos  x } }{ { \sin  x } }  } }{ x }$$

$$= \mathop {\lim }\limits_{x \to 0}$$ $$\dfrac { { \sin  x\cos  x-\cos  x } }{ { x\sin  x } }$$

$$= \mathop {\lim }\limits_{x \to 0}$$ $$\dfrac{{\cos x\left( {\sin x - 1} \right)}}{{x\sin x}}$$

Now , 

$$\mathop {\lim }\limits_{x \to {0^ - }}$$ $$\dfrac { { \cos  x\left( { \sin  x-1 } \right)  } }{ { x\sin  x } }$$

$$\, = \,\,\mathop {\lim }\limits_{h \to 0}$$ $$\dfrac { { \cos  \left( { 0-h } \right) \left( { \sin  \left( { 0-h } \right) -1 } \right)  } }{ { \left( { 0-h } \right) \sin  \left( { 0-h } \right)  } }$$

$$= \,\mathop {\lim }\limits_{h \to 0}$$ $$\dfrac{{\cosh \left( { - \sinh  - 1} \right)}}{{ - h \times  - \sinh }}$$

$$= \,\,\mathop {\lim }\limits_{h \to 0}$$ $$\dfrac { { -\cosh  \left( { \sinh  +1 } \right)  } }{ { h\sinh   } }$$

$$=-\infty$$

$$\mathop {\lim }\limits_{x \to {0^ + }}$$ $$\dfrac { { \cos  x\left( { \sin  x-1 } \right)  } }{ { x\sin  x } }$$

$$= \,\,\mathop {\lim }\limits_{h \to 0}$$ $$\dfrac { { \cos  \left( { 0+h } \right) \left( { \sin  \left( { 0+h } \right) -1 } \right)  } }{ { \left( { 0+h } \right) \sin  \left( { 0+h } \right)  } }$$

$$= \,\mathop {\lim }\limits_{h \to 0}$$ $$\dfrac { { \cosh  \left( { \sinh  -1 } \right)  } }{ { h\sinh   } }$$

$$\,\,\,\,\, = \dfrac{{1 \times  - 1}}{0}$$

$$=-\infty$$

So , $$\mathop {\lim }\limits_{x \to 0}$$ $$\dfrac { { \cos  x-\cot  x } }{ x } \, =\, -\infty$$

$${\lim}_{x\rightarrow 0}{\left({\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}^{\frac{1}{4}}-x\right)}$$

We have,

$$\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\left( x\sin \left( \dfrac{3}{x} \right) \right)$$

$$\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\left( 3\dfrac{\sin \left( \dfrac{3}{x} \right)}{\left( \dfrac{3}{x} \right)} \right)$$

Let $$h=\dfrac{3}{x}$$

Therefore,

$$\displaystyle \underset{h\to 0}{\mathop{\lim }}\,\left( 3\dfrac{\sinh }{h} \right)$$

$$\displaystyle 3\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{\sinh }{h} \right)$$

$$=3\times 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \,\because \underset{x\to 0}{\mathop{\lim }}\,\left( \dfrac{\sin x}{x} \right) \right]$$

$$=3$$

Hence, this is the answer.

Consider the given function:

  $$f\left( x \right)=\dfrac{{{x}^{15}}-1}{{{x}^{10}}-1}$$

  $$=\dfrac{{{\left( {{x}^{5}} \right)}^{3}}-{{\left( 1 \right)}^{3}}}{{{\left( x \right)}^{2}}-{{\left( 1 \right)}^{2}}}$$

  $$=\dfrac{\left( {{x}^{5}}-1 \right)\left( {{x}^{10}}+1+{{x}^{5}} \right)}{\left( {{x}^{5}}-1 \right)\left( {{x}^{5}}+1 \right)}$$

  $$=\dfrac{{{x}^{10}}+{{x}^{5}}+1}{\left( {{x}^{5}}+1 \right)}$$

  $$=1$$

Hence this is the answer.

$$\displaystyle \lim_{x\rightarrow 0}((x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}})$$

$$\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 + \sin x}  - \sqrt {1 - \sin x} }}{x}$$

$$\underset{n \rightarrow \infty}{lim} \left(1 + \dfrac{4}{n} \right)^{3n}$$

$$\displaystyle \lim_{x\rightarrow 1} \dfrac {(ln (1 + x) - ln 2)(3 . 4^{x - 1} - 3x)}{[(7 + x)^{\dfrac {1}{3}} - (1 + 3x)^{\dfrac {1}{2}}] \sin (x - 1)}$$

$$\displaystyle \lim_{x\rightarrow \pi/2} \dfrac {1 + \cos 2x}{(\pi - 2x)^{2}}$$

$${ \lim   }_{ x\to a }\cfrac { { 2x-\sqrt { { x^{ 2 } }+3{ a^{ 2 } } }  } }{ { \sqrt { x+a } -\sqrt { 2a }  } } =\sqrt { 2 }$$

We have,

$$\lim _{ x\longrightarrow 2 }{ \cfrac { { x }^{ 7 }-128 }{ { x }^{ 5 }-32 }  } \\ =\lim _{ x\longrightarrow 2 }{ \cfrac { { x }^{ 7 }-{ 2 }^{ 7 } }{ { x }^{ 5 }-{ 2 }^{ 5 } }  } =\lim _{ x\longrightarrow 2 }{ \cfrac { \cfrac { { x }^{ 7 }-{ 2 }^{ 7 } }{ x-2 }  }{ \cfrac { { x }^{ 5 }-{ 2 }^{ 5 } }{ x-2 }  }  } \\ =\cfrac { \lim _{ x\longrightarrow 2 }{ \cfrac { { x }^{ 7 }-{ 2 }^{ 7 } }{ { x }-2 }  }  }{ \lim _{ x\longrightarrow 2 }{ \cfrac { { x }^{ 5 }-{ 2 }^{ 5 } }{ { x }-2 }  }  } \\ =\cfrac { 7.{ 2 }^{ 6 } }{ 5.{ 2 }^{ 4 } } \\ =\cfrac { 7 }{ 5 } \times { 2 }^{ 2 }=\cfrac { 7 }{ 5 } \times 4=\cfrac { 28 }{ 5 }$$

$$\cos x$$ at $$x\rightarrow \infty$$ lies between $$-1$$ to $$1.$$

We have,

$$={\lim}_{x\rightarrow 0}{\dfrac{\dfrac{{\sin}^{2}{2x}}{{\left(2x\right)}^{2}}\times4{x}^{2}}{\dfrac{{\sin}^{2}{4x}}{{\left(4x\right)}^{2}}\times 16{x}^{2}}}$$

$$\displaystyle\lim_{x\to 3} \left(\dfrac{x^4-81}{2x^2-5x-3}\right)$$

We have,

$$\displaystyle \lim_{x\to 0^+}\dfrac {2}{x^{1/5}}=a$$

$$\displaystyle \lim_{x\rightarrow 0^+}{\dfrac{1}{3x}}=a$$

$$\underset{x \rightarrow 1}{\lim} \dfrac{x^2 + 1}{x + 1}$$

$$\displaystyle \lim_{x\rightarrow \tfrac{-5}{2}}[x]$$

Let L =  $$\dfrac{ 1 -x + \log x}{1 - \sqrt{2x - x^2}}$$

$$\displaystyle \lim _{x \rightarrow 0}\left(\dfrac{1+5 x^{2}}{1+3 x^{2}}\right)^{1 / x^{2}}$$
$$=\dfrac{\displaystyle \lim _{x \rightarrow 0}\left(1+5 x^{2}\right)^{1 / x^{2}}}{\lim \left(1+3 x^{2}\right)^{1 / x^2}}$$
$$\dfrac{\displaystyle \lim _{x \rightarrow 0}\left\{\left(1+5 x^{2}\right)^{\dfrac{1}{5 x^{2}}}\right\}^{5}}{\displaystyle \lim _{x \rightarrow 0}\left\{\left(1+3 x^{2}\right)^{\dfrac{1}{3 x^{2}}}\right\}^{3}}$$
$$= e^{5-3}=e^{2}$$

$$\operatorname{Let} L= \displaystyle lim_{x \rightarrow \infty} \dfrac{\log _{e}\left(\log _{e} x\right)}{e^{\sqrt{x}}}=\left(\dfrac{\infty}{\infty} \text { form }\right)$$
$$=\displaystyle \lim _{x \rightarrow \infty} \dfrac{\dfrac{1}{x \log _{e} x}}{e^{\sqrt{x}} \dfrac{1}{2 \sqrt{x}}}$$
$$=\displaystyle \lim _{x \rightarrow \infty} \dfrac{2 \sqrt{x}}{e^{\sqrt{x}} x \log _{e} x}$$
$$=\displaystyle \lim _{x \rightarrow \infty} \dfrac{2}{e^{\sqrt{x}} \sqrt{x} \log _{e} x}$$
$$=0$$

(A) For a=0, $$I=\displaystyle \int_{0}^{T}sin^2xdx= I=\displaystyle \int_{0}^{T} \dfrac{1-cos2x}{2}dx$$
$$=\left[\dfrac{x}{2}-\dfrac{1}{4}sin2x\right]_{0}^{T} =\dfrac{T}{2}-\dfrac{1}{4}sin2T$$
$$\left[x-\dfrac{1}{4}sin2x-\dfrac{1}{4a}sin2ax\dfrac{sin(a-1)x}{a-1}-dfrac{sin(a+1)x}{a+1}\right]_{0}^{T}$$
$$\therefore L=\dfrac{1}{2}-\left[\dfrac{1}{4}-\displaystyle \lim_{T→∞}\dfrac{1}{4}\dfrac{sin2T}{T}\right]=\dfrac{1}{2}$$

(B) For a=1, $$\displaystyle \int_{0}^{T}4sin^2xdx \Rightarrow L=2$$

(C) For a=-1, $$\displaystyle \int_{0}^{T} 0dx=0 \Rightarrow L=0$$

(D) For $$a\neq 0,-1,1,$$
$$I=\displaystyle \int_{0}^{T}(sin^2x+sin^2ax +2sinx.sinax)dx$$
$$= \displaystyle \int_{0}^{T} \left( \dfrac{1-cos2x}{2}+\dfrac{1-cos2ax}{2}+cos(a-1)x-cos(a+1)x\right)dx$$
$$=\left[ x-\dfrac{1}{4}sin2x-\dfrac{1}{4a}sin2ax\dfrac{sin(a-1)x}{a-1}-\dfrac{sin(a+1)x}{a+1}\right]_{0}^{T}$$
$$\therefore L=\displaystyle \lim_{T→∞}\dfrac{T}{T}-\lim_{T→∞}\dfrac{1}{T}$$
$$\left[\dfrac{1}{4}sin2x-\dfrac{1}{4a}sin2ax+\dfrac{sin(a-1)x}{a-1}-\dfrac{sin(a+1)x}{a+1}\right]_{0}^{T}$$
$$\Rightarrow L=1$$

$$\displaystyle \lim _{x \rightarrow-\infty}\left[\dfrac{x^{4} \sin \left(\dfrac{1}{x}\right)+x^{2}}{\left(1+|x|^{3}\right)}\right]$$
$$=\displaystyle \lim _{x \rightarrow-\infty}\left[\dfrac{x \sin \left(\dfrac{1}{x}\right)+\dfrac{1}{x}}{\dfrac{1}{x^{3}}-1}\right]$$
$$=\displaystyle \lim _{x \rightarrow-\infty}\left[\dfrac{\dfrac{\sin \left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}+\dfrac{1}{x}}{\dfrac{1}{x^{3}}-1}\right]=\dfrac{1+0}{0-1}=-1$$

$$\displaystyle \lim_{x\rightarrow \infty} 2^{x} \sin \left (\dfrac {a}{2^{x}}\right )$$
$$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {a\times \sin \left (\dfrac {a}{2^{x}}\right )}{a} \times 2^{x}$$
$$= \displaystyle \lim_{x\rightarrow \infty} \dfrac {a\sin \left (\dfrac {a}{2^{x}}\right )}{\left (\dfrac {a}{2^{x}}\right )}$$
$$= a\times \displaystyle \lim_{x\rightarrow \infty} \dfrac {\sin \left (\dfrac {a}{2^{x}}\right )}{\left (\dfrac {a}{2^{x}}\right )}$$
$$= a \times 1 = a$$ .

$$\displaystyle \lim_{x\rightarrow 1}\dfrac {e^{x} - e^{-x}}{e^{x} + e^{-x}}$$
$$= \displaystyle \lim_{x\rightarrow 1} e^{-x} \left [\dfrac {e^{x}}{e^{-x}} - 1\right ] / e^{-x} \left [\dfrac {e^{x}}{e^{-x}} + 1\right ]$$
$$= \displaystyle \lim_{x\rightarrow 1} \dfrac {e^{x + x} - 1}{e^{x + x} + 1}$$
$$= \displaystyle \lim_{x\rightarrow 1} \dfrac {e^{2x} - 1}{e^{2x} + 1}$$
$$= \dfrac {e^{2(1)} - 1}{e^{2(1)} + 1} = \dfrac {e^{2} - 1}{e^{2} + 1}$$ .

$$\displaystyle \lim_{x\rightarrow 0} \dfrac {ae^{x} - b \cos x + ce^{-x}}{x\sin x} = 2 .... (i)$$
$$\Rightarrow \displaystyle \lim_{x\rightarrow 0} \dfrac {ae^{x} + b\sin x - ce^{-x}}{x\cos x + \sin x} = 2$$
$$\Rightarrow \displaystyle \lim_{x\rightarrow 0} \dfrac {ae^{x} + b\cos x + ce^{-x}}{-x\sin x + \cos x + \cos x} = 2$$
$$\Rightarrow \dfrac {ae^{0} + b\cos 0 + ce^{0}}{0 + \cos 0 + \cos 0} = 2$$
$$\Rightarrow \dfrac {a + b + c}{2} = 2 .... (ii)$$
From equation (i),
$$\displaystyle \lim_{x\rightarrow 0} \dfrac {a \left (1 + x + \dfrac {x^{2}}{2!} + ...\right ) - b \left (1 - \dfrac {x^{2}}{2!} + \dfrac {x^{4}}{4!} - ... \right ) + c\left (1 - x + \dfrac {x^{2}}{2!} - ...\right )}{x\left (x - \dfrac {x^{3}}{3!} + \dfrac {x^{5}}{5!} - ...\right )} = 2$$
$$\Rightarrow \displaystyle \lim_{x\rightarrow 0} \dfrac {(a - b + c) + x(a - c) + x^{2} \left (\dfrac {a}{2} + \dfrac {b}{2} + \dfrac {c}{2}\right ) + ...}{x^{2} \left (1 - \dfrac {x^{2}}{3!} + ...\right )} = 2$$
$$a = c, c - b + c = 0, a + b + c = 4$$
$$\Rightarrow 2a - b = 0$$ ,
$$\Rightarrow b = 2a \Rightarrow a = 1 = c, b = 2$$
Hence, $$a = 1, b = 2, c = 1$$ .

Given  $$\displaystyle { f }({ x })=\frac { 1-\cos ^{ 2 }{ x } +\sin ^{ 2 }{ x }  }{ \sqrt { { x }^{ 2 }+1 } -1 } ({ x }\neq 0)$$
Since, $$f(x)$$ is continuous at $$x=0$$
$$\lim _{ x\rightarrow 0 }{ f(x)=f(0) }$$

$$\lim_{x\rightarrow 0}\displaystyle \frac{1-cos^{2}x+sin^{2}x}{\sqrt{x^{2}+1}-1}$$

$$\displaystyle =\lim_{x\rightarrow 0}\frac{2sin^{2}x}{x^{2}}(\sqrt{x^{2}+1}+1)$$

$$=2\lim_{x\rightarrow 0}\left (\displaystyle  \frac{sin\, x}{x} \right )^2\lim_{x\rightarrow 0}(\sqrt{x^{2}+1}+1)$$
$$=2 \times 1 \times 2=4$$
$$\therefore f(0)=4$$  

$$\displaystyle \lim_{x\rightarrow 0}\frac{a\left [ x-\displaystyle \frac{x^3}{6}+\frac{x^5}{120}-....... \right ]-bx +cx^2+ x^3}{2x^2\left [x-\displaystyle \frac{x^2}{2}+\frac{x^3}{3}.........  \right ]-2x^3+x^4}$$
$$=\lim_{x\rightarrow 0}\displaystyle \frac{(a- b) x +cx^2+\left [ 1-\displaystyle \frac{a}{6} \right ]x^3+\displaystyle \frac{ax^5}{120}+.....}{\displaystyle \frac{2x^5}{3}-\displaystyle \frac{x^6}{2}+......}$$
for this limit to be exist, $$a- b = 0, c = 0$$ , & $$1-\displaystyle \frac{a}{6} =0$$
$$\Rightarrow a = b = 6$$ & $$c = 0$$ .
then $$f(0) =\displaystyle \frac{6}{120}.\frac{3}{2}=\frac{3}{40}  \Rightarrow   200\: f(0) = 15$$

Using $$C_{3}\rightarrow C_{3}-C_{2}$$ and $$C_{2}\rightarrow C_{2}-C_{1}$$ , we can write
$$\displaystyle \frac{\Delta }{h^{2}}=\begin{vmatrix} \tan x & [\tan (x+h)-\tan x]/h & [\tan (x+2h)-\tan (x+h)]/h \\ \tan (x+2h) & [\tan x-\tan (x+2h)]/h & [\tan (x+h)-\tan x]/h  \\ \tan (x+h) & [\tan (x+2h)-\tan (x+h)]/h & [\tan x-\tan (x+2h)]/h \end{vmatrix}$$
But 
$$\displaystyle \lim_{h\rightarrow 0}\frac{\tan (x+h)-\tan x}{h}$$
$$=\displaystyle \lim_{h\rightarrow 0}\frac{1}{h}\frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h)\cos x}$$
$$=\displaystyle \lim_{h\rightarrow 0}\frac{1}{h}\frac{\sin h}{\cos (x+h)\cos x}=\sec ^{2}x$$ or (use $$\frac{\mathrm{d} }{\mathrm{d} x}(\tan x)=\sec ^{2}x$$ )
Similarly, $$=\displaystyle \lim_{h\rightarrow 0}\frac{\tan (x+2h)-\tan (x+h)}{h}=\sec ^{2}x$$
and $$=\displaystyle \lim_{h\rightarrow 0}\frac{\tan x-\tan (x+2h)}{h}=-2\sec ^{2}x$$
Thus,

$$\displaystyle \frac{\Delta (x)}{h^{2}}=\begin{vmatrix} \tan x  & \sec ^{2}x  & \sec ^{2}x \\ \tan x & -\sec ^{2}x & \sec ^{2}x  \\ \tan x & \sec ^{2}x  & -2\sec ^{2}x \end{vmatrix}$$   $$=\displaystyle \begin{vmatrix} \tan x  & \sec ^{2}x  & \sec ^{2}x \\ 0 & -3\sec ^{2}x & 0  \\ 0 & 0  & -3\sec ^ {2}x \end{vmatrix}$$
$$=\displaystyle 9\:\tan x\:\sec ^{4}x$$ [Using $$R_{2}\rightarrow R_{2}-R_{1}$$  and $$R_{3}\rightarrow R_{3}-R_{1}$$ ]
So $$\displaystyle \lim_{h\rightarrow 0}\frac{\sqrt{3}\Delta (\pi /3)}{h^{2}}=9\times 3\times 2^{4}=216$$

$$\displaystyle\lim\limits_{n\to\infty} \dfrac{5^{n+1}+3^n-2^{2n}}{5^n+2^n+3^{n+2}}$$ $$=\dfrac{5+0-0}{1+0+0}$$

$$\displaystyle\lim\limits_{x\to 0} \dfrac{\cos 4x-\cos 6x}{\sin^25x}$$