\(\frac{1}{2}cot x\sqrt{sinx}\)

2

We have \(\overset{lim}{x\rightarrow 0^{-}}\)f(x)=\(\overset{lim}{h\rightarrow 0}\)f(0-h)=\(\overset{lim}{h\rightarrow 0}\)f(-h)=\(\overset{lim}{h\rightarrow 0}\)f(h)
[∵ f is an even function so f(- h)=f(h)]
=\(\overset{lim}{h\rightarrow 0}\)f(0+h)=\(\overset{lim}{x\rightarrow 0^{+}}\)f(x)
Thus \(\overset{lim}{x\rightarrow 0^{-}}\)f(x)=\(\overset{lim}{x\rightarrow 0^{+}}\)f(x) when f is an even function.

\(\frac{1}{2}\)

\(\lim_ { x\rightarrow 0 }\frac { { e }^{ x }-1-x-{ x }^{ 2 } }{ { x }^{ 2 } } \)
\(\Rightarrow \lim_ { x\rightarrow 0 } \left[ \frac { \left( 1+x+\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } ... \right) -1-x }{ { x }^{ 2 } } -1 \right] \left[ \because { e }^{ x }=1+x+\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } ... \right] \)
\(\Rightarrow \lim_ { x\rightarrow 0 } \frac { \left( \frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } +.. \right) }{ { x }^{ 2 } } -1\)
\(\Rightarrow \lim_ { x\rightarrow 0 } \frac { { x }^{ 2 }\left( \frac { 1 }{ 2! } +\frac { { x } }{ 3! } +.... \right) }{ { x }^{ 2 } } -1\)
\(\Rightarrow \lim_{ x\rightarrow 0 } \frac { 1 }{ 2! } +\frac { { x } }{ 3! } -1\Rightarrow \frac { 1 }{ 2 } -1\Rightarrow -\frac { 1 }{ 2 } \)

Here \(\lim_{x \rightarrow 2} \frac{x^{2}-5x+6}{x^{2}-4}\)\([\frac{0}{0} form]\)
=\(\frac{(x-2)(x-3)}{(x+2)(x-2)}=\lim_{x\rightarrow 2} \frac{x-3}{x+2}=​​\frac{2-3}{2+2}=\frac{1}{4}\)

Let \(f\left( x \right) =\cos { \left( { x }^{ 2 }+1 \right) } \)
We know by first principle,
\(f'\left( x \right) =\lim _{ h\rightarrow 0 }{ \frac { f\left( x+h \right) -f\left( x \right) }{ h } } \)
\(\therefore f'\left( x \right) =\lim _{ h\rightarrow 0 }{ \frac { \cos { \left[ \left( x+h \right) ^{ 2 }+1 \right] -\cos { \left( { x }^{ 2 }+1 \right) } } }{ h } }\)|
\(\left[ \therefore f\left( x \right) =\cos { \left( { x }^{ 2 }+1 \right) } \right] \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -2\sin { \frac { \left( x+h \right) ^{ 2 }+1+{ x }^{ 2 }+1 }{ 2 } \sin { \frac { \left( x+h \right) ^{ 2 }+1-\left( { x }^{ 2 }+1 \right) }{ 2 } } } }{ h } } \)
\( \left[ \therefore \cos { C } -\cos { D=-2\sin { \left( \frac { C+D }{ 2 } \right) } \sin { \left( \frac { C-D }{ 2 } \right) } } \right] \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -2\sin { \frac { \left( x+h \right) ^{ 2 }+{ x }^{ 2 }+2 }{ 2 } \sin { \frac { \left( x+h \right) ^{ 2 }-{ x }^{ 2 } }{ 2 } } } }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { \left\{ -2\sin { \frac { \left( x+h \right) ^{ 2 }+{ x }^{ 2 }+2 }{ 2 } \left[ \sin { \frac { \left( x+h \right) ^{ 2 }-{ x }^{ 2 } }{ 2 } } \right] } \left[ \frac { \left( x+h \right) ^{ 2 }-{ x }^{ 2 } }{ 2 } \right] \right\} }{ h\times \frac { \left( x+h \right) ^{ 2 }-{ x }^{ 2 } }{ 2 } } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -2\sin { \frac { \left( x+h \right) ^{ 2 }+{ x }^{ 2 }+2 }{ 2 } \times \frac { { x }^{ 2 }+{ h }^{ 2 }+2xh-{ x }^{ 2 } }{ 2 } } }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -2\sin { \left[ \frac { \left( x+h \right) ^{ 2 }+{ x }^{ 2 }+2 }{ 2 } \right] } \left( 2x+h \right) \frac { h }{ 2 } }{ h } } \)
\(=-2\sin { \left( \frac { 2{ x }^{ 2 }+2 }{ 2 } \right) } \times 2x\times \frac { 1 }{ 2 } =-2x\sin { \left( { x }^{ 2 }+1 \right) } \)

\( \lim _{x \rightarrow \pi / 6} \frac{\cot ^{2} x-3}{\operatorname{cosec} x-2} \)
\(= \lim _{x \rightarrow \pi / 6} \frac{\operatorname{cosec}^{2} x-1-3}{\operatorname{cosec} x-2} \quad\left[\because \operatorname{cosec}^{2} x-\cot ^{2} x=1\right] \)
\(= \lim _{x \rightarrow \pi / 6} \frac{\operatorname{cosec}^{2} x-4}{\operatorname{cosec} x-2} \)
\(= \lim _{x \rightarrow \pi / 6} \frac{(\operatorname{cosec} x-2)(\operatorname{cosec} x+2)}{(\operatorname{cosec} x-2)} \quad[\text { by factorisation }] \)
\(=\lim _{x \rightarrow \pi / 6}(\operatorname{cosec} x+2)\)
\(= \operatorname{cosec} \frac{\pi}{6}+2=2+2=4\)

Put 1+x=y, then y \(\rightarrow\) 1 as x \(\rightarrow\) 0.
Now,  \(\lim_ { x\rightarrow 0 }{ lim } \frac { { \left( 1+x \right) }^{ n }-1 }{ x } =\lim_ { y\rightarrow 1 }{ lim } \frac { { y }^{ n }-1 }{ y-1 } =\lim_ { y\rightarrow 1 }{ lim } \frac { { y }^{ n }-{ 1 }^{ n } }{ y-1 } \)
n

\(f^{\prime}(100)=\lim _{h \rightarrow 0} \frac{f(100+h)-f(100)}{h} \)
\(=\lim _{h \rightarrow 0} \frac{99(100+h)-99(100)}{h} \)
\(=\lim _{h \rightarrow 0} \frac{99 \times 100+99 h-99 \times 100}{h} \)
\(=\lim _{h \rightarrow 0} \frac{99 h}{h} \)
\(=\lim _{h \rightarrow 0}(99)=99\)
\(\text { Thus, the derivative of } 99 x \text { at } x=100 \text { is } 99 \text { . }\)

\(\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{x^{2}} \)
\(=\lim _{x \rightarrow 0} \frac{e^{2 x}+1-2 e^{x}}{x^{2} e^{x}} \)
\(=\lim _{x \rightarrow 0}\left(\frac{e^{x}-1}{x}\right)^{2} \times e^{-x} \)
\(=\lim _{x \rightarrow 0}\left(\frac{e^{x}-1}{x}\right)^{2} \times \lim _{x \rightarrow 0} e^{-x}=(1)^{2} \times e^{0}=1\)

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

Given functions are f(x) = \({ x }^{ 2 }+4\)and g(x) = x-3
Clearly, \(\lim _{ x\rightarrow 5 }{ f(x) } =\lim _{ x\rightarrow 5 }{ { x }^{ 2 }+4 } \)
\(={ (5) }^{ 2 }+4=25+4=29 ...(i)\)
and \(\lim _{ x\rightarrow 5 }{ \quad g(x) } =\lim _{ x\rightarrow 5 }{ (x-3)=(5-3)=2 } ...(ii)\)
(ii) \(\lim _{ x\rightarrow 5 }{ \quad [f(x) } -g(x)]= \lim _{ x\rightarrow 5 }{ f(x) } -\lim _{ x\rightarrow 5 }{ g(x) } \)
=29-2 [from Eqs.(i) and (ii)]
=27

Given \(\lim _{ x\rightarrow 4 }{ \frac { |x-4| }{ x-4 } } \)
\(LHL=\lim _{ { X\rightarrow 4 }^{ 1 } }{ \frac { -(x-4) }{ x-4 } } =1 \quad \left[ \because |x-4|=-(x-4),x<4 \right] \)
\(RHL=\lim _{ { X\rightarrow 4 }^{ + } }{ \frac { (x-4) }{ x-4 } } =1\quad \left[ \because |x-4|=(x-4),x<4 \right] \)

Put x=e+h
then \(x\rightarrow e \Rightarrow h\rightarrow 0\)
ஃ \(\overset{Lt}{h\rightarrow 0}\frac{log x-1}{x-e}=\overset{Lt}{h\rightarrow 0}\frac{log(e+h)-log e}{e+h-e}\)   [∵ log e=1]
=\(\overset{Lt}{h\rightarrow 0} \frac{log[\frac{e+h}{e}]}{h} \Rightarrow \overset{Lt}{h\rightarrow 0}\frac{log[1+\frac{h}{e}]}{\frac{h}{e}\times e}\)
∴ \(\frac{h}{e}\rightarrow 0 \Rightarrow \frac{1}{e}.\frac{1}{1} \Rightarrow \frac{1}{e} \)
∴ \(\overset{Lt}{x\rightarrow e}\frac{log x-1}{x-e}=\frac{1}{e}\).

\(\text { The given function is }\)
\(f(x)=\left\{\begin{array}{ll} a+b x, & x<1 \\ 4, & x=1 \\ b-a x & x>1 \end{array}\right.\)
\(\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(a+b x)=a+b \)
\(\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(b-a x)=b-a \)
\(f(1)=4\)
\(\text { It is given that } \lim _{x \rightarrow 1} f(x)=f(1) \text { . }\)
\(\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x)=f(1) \)
\(\Rightarrow a+b=4 \text { and } b-a=4\)
\(\text { On solving these two equations, we obtain } a=0 \text { and } b=4 \text { . }\)
\(\text { Thus, the respective possible values of } a \text { and } b \text { are } 0 \text { and } 4\)

\(\text { Let } f(x)=x^{4}(5 \sin x-3 \cos x)\)
\(\text { By product rule, }\)
\(f^{\prime}(x) =x^{4} \frac{d}{d x}(5 \sin x-3 \cos x)+(5 \sin x-3 \cos x) \frac{d}{d x}\left(x^{4}\right) \)
\(=x^{4}\left[5 \frac{d}{d x}(\sin x)-3 \frac{d}{d x}(\cos x)\right]+(5 \sin x-3 \cos x) \frac{d}{d x}\left(x^{4}\right) \)
\(=x^{4}[5 \cos x-3(-\sin x)]+(5 \sin x-3 \cos x)\left(4 x^{3}\right) \)
\(=x^{3}[5 x \cos x+3 x \sin x+20 \sin x-12 \cos x]\)