\(\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x}=\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x} \times \frac{\sin x}{\sin x}\)
[ Multiplying numerator and denominator by sin x]
\(=\begin{matrix} lim \\ x\rightarrow 0 \end{matrix}\left[ \frac { { e }^{ \sin { x } }-1 }{ { \sin { x } } } \times \frac { \sin { x } }{ x } \right] =\begin{matrix} lim \\ x\rightarrow 0 \end{matrix}\frac { { e }^{ \sin { x } }-1 }{ { \sin { x } } } \times \begin{matrix} lim \\ x\rightarrow 0 \end{matrix}\frac { \sin { x } }{ x } \)
\(=1\times 1=1 \quad \left[ \because \begin{matrix} lim \\ \theta \rightarrow 0 \end{matrix}\frac { { e }^{ \theta }-1 }{ { \theta } } =1\quad and\begin{matrix} lim \\ \theta \rightarrow 0 \end{matrix}\frac { \sin { \theta } }{ { \theta } } =1\quad \right] \)

We have, \(y=\sqrt { \frac { 1-cos2x }{ 1+cos2x } } =\sqrt { \frac { 2{ sin }^{ 2 }x }{ 2{ cos }^{ 2 }x } } =\sqrt { { tan }^{ 2 }x } \)
\( [\because cos2\theta =1-2{ sin }^{ 2 }\theta =1-cos2\theta \quad and\quad cos2\theta =2{ cos }^{ 2 }\theta -1\)
\(\Rightarrow 2{ cos }^{ 2 }\theta =1+cos2\theta ]\)
\(\Rightarrow y=\left| tanx \right| ,\quad where\quad x\in (0,\frac { \pi }{ 2 } )\cup (\frac { \pi }{ 2 } ,\pi )\)
\(\text { Now, } \quad y=\left\{\begin{array}{ll} \tan x, x \in\left(0, \frac{\pi}{2}\right) \\ -\tan x, x \in\left(\frac{\pi}{2}, \pi\right) \end{array}\right.\)
\(\therefore \frac{d y}{d x}=\left\{\begin{aligned} \sec ^{2} x, \text { if } x \in\left(0, \frac{\pi}{2}\right) \\ -\sec ^{2} x, \text { if } x \in\left(\frac{\pi}{2}, \pi\right) \end{aligned}\right.\)

Given limit = \(\lim_ { x\rightarrow 1 }{ lim } \frac { \frac { { x }^{ m }-1 }{ x-1 } }{ \frac { { x }^{ n }-1 }{ x-1 } } =\frac { \underset { x\rightarrow 1 }{ lim } \frac { { x }^{ m }-{ 1 }^{ m } }{ x-1 } }{ \underset { x\rightarrow 1 }{ lim } \frac { { x }^{ n }-{ 1 }^{ n } }{ x-1 } } =\frac { m }{ n } \)

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

\(Given\ limit=\lim_ { x\rightarrow 0 }{ lim } \frac { { x }^{ 2 }cosx }{ 2{ sin }^{ 2 }\frac { x }{ 2 } } \lim_ { x\rightarrow 0 }{ 2lim } \frac { { \left( \frac { x }{ 2 } \right) }^{ 2 } }{ { sin }^{ 2 }\frac { x }{ 2 } } .\lim_ { x\rightarrow 0 }{ lim } cosx\)
2

\(\lim_ { x\rightarrow 0 }{ lim } \frac { tan2x-x }{ 3x-sinx } =\lim_ { x\rightarrow 0 }{ lim } \frac { x\left[ \frac { tan2x }{ x } -1 \right] }{ x\left[ 3-\frac { sinx }{ x } \right] } =\frac { \lim_ { x\rightarrow 0 }{ lim } 2\times \frac { tan2x }{ 2x } -1 }{ 3-\lim_ { x\rightarrow 0 }{ lim } \frac { sinx }{ x } } \)
\(\frac { 1 }{ 2 } \)

Given limit = \(\lim_ { x\rightarrow 0 }{ lim } \left[ \frac { tanx }{ x } -\frac { sinx }{ x } \right] \)
=0

\(\lim _{ x\rightarrow 3 }{ { (4x }^{ 3 }-{ 2x }^{ 2 } } -x+1)\)
\(=4\lim _{ x\rightarrow 3 }{ { x }^{ 3 }-2\lim _{ x\rightarrow 3 }{ { x }^{ 2 } } } -\lim _{ x\rightarrow 3 }{ x } +\lim _{ x\rightarrow 3 }{ 1 } \)
\(={ 4(3) }^{ 3 }-{ 2(3) }^{ 2 }-3+1=108-18-2=88\)

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 \(\lim _{ x\rightarrow 5 }{ \quad g(x) } =\lim _{ x\rightarrow 5 }{ (x-3)=(5-3)=2 } ...(ii)\)
(iii) \(\lim _{ x\rightarrow 5 }{ \quad [f(x) } \times g(x)]= \lim _{ x\rightarrow 5 }{ f(x) } \times \lim _{ x\rightarrow 5 }{ g(x) } \)
=29 x 2 =58     [from Eqs. (i) and (ii)]

Here f(x)=\(\frac { { (px }^{ 2 }+qx+r) }{ ax+b } \)
\(\therefore f(x)=\frac { d }{ dx } \left[ \frac { { px }^{ 2 }+qx+r }{ ax+b } \right] \)
\( =\frac { (ax+b)\frac { d }{ dx } ({ px }^{ 2 }+qx+r)-({ px }^{ 2 }+qx+r)\frac { d }{ dx } (ax+b) }{ { (ax+b) }^{ 2 } } \)
\( =\frac { (ax+b)(2px+q)-({ px }^{ 2 }+qx+r)(a) }{ { (ax+b) }^{ 2 } } \)
\(=\frac { { 2px }^{ 2 }+aqx+2bpx+bq-{ apx }^{ 2 }-apx-ar }{ { (ax+b) }^{ 2 } } \)
\(=\frac { { apx }^{ 2 }+2bpx+bq-ar }{ { (ax+b) }^{ 2 } } \)

\({ f }^{ ' }(x)=\lim_ { h\rightarrow 0 }{ lim } \frac { cot(x=h)-cotx }{ h }\)
\(\Rightarrow { f }^{ ' }(x)= \lim_ { h\rightarrow 0 }{ lim } \frac { sinx\quad cos(x+h)-sin(x+h)cosx }{ hsin(x+h)sinx } \)
\(=\lim_{ h\rightarrow 0 }{ lim } \frac { sin(-h) }{ hsin(x+h)sinx } \)
\(-cose{ c }^{ 2 }x\)

\(Given\ limit=\lim_ { x\rightarrow 0 }{ lim } \frac { cos2x-1 }{ cosx-1 } \)
\( =\lim_ { x\rightarrow 0 }{ lim } \frac { 2sin^{ 2 }x }{ 2sin^{ 2 }\frac { x }{ 2 } } =\lim_ { x\rightarrow 0 }{ lim } { \left( \frac { sinx }{ x } \right) }^{ 2 }\times { \left( \frac { \frac { x }{ 2 } }{ sin\frac { x }{ 2 } } \right) }^{ 2 }\times 4\)
4

\(LHL=\lim _{ x\rightarrow { 1 }^{ - } }{ a+bx)= } \lim _{ h\rightarrow { 0 } }{ [a+b(1-h)]=a+b } \)
\(RHL=\lim _{ x\rightarrow { 1 }^{ + } }{ (b-ax) } =\lim _{ h\rightarrow { 0 } }{ [b-a(1+h)]=b-a } \)
\(\because LHL=RHL=f\left( 1 \right) \Rightarrow A+B=B-A=4\)
Ans. A = 0, B = 4

Here f(x)=\(\frac { a+b\quad sin\quad x }{ c+d\quad cos\quad x } \)
\(\therefore f'(x)=\frac { d }{ dx } \left[ \frac { a+b\quad sin\quad x }{ c+d\quad cos\quad x } \right] \)
= \(\frac { (c+dcosx)\frac { d }{ dx } (a+b\quad sinx)-(a+b\quad sinx)\frac { d }{ dx } (c+d\quad cosx) }{ c+d\quad cosx^{ 2 } } \)
\(=\frac { (a-b\quad sinx)(-d\quad sinx) }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bc\quad cosx+bd\quad cos^{ 2 }x+\quad ad\quad sinx+bd\quad sin^{ 2 }x }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bc\quad cosx+bd\quad sinx+\quad bd(cos^{ 2 }x+sin^{ 2 }x) }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bd(cos^{ 2 }x+sin^{ 2 }x) }{ (c+d\quad cosx)^{ 2 } } \)
\(=\frac { bc\quad cosx+ad\quad sinx+bd }{ (c+d\quad cosx)^{ 2 } } \)

Here f(x)=\(\frac { sin\quad (x+a) }{ cos\quad x } \)
\(\therefore \quad f'(x)=\frac { d }{ dx } \left[ \frac { sin(x+a) }{ cos\quad x } \right] \)
= \(\frac { cos\quad x\frac { d }{ dx } [sin(x+a)]-sin(x+a)\frac { d }{ dx } (cos\quad x) }{ cos^{ 2 }x } \)
\(=\frac { cos\quad x.cos(x+a)-sin(x+a)(-sin\quad x) }{ cos^{ 2 } } \)
\(=\frac { cos\quad x.cos(x+a)+sin\quad x\quad sin(x+a) }{ cos^{ 2 }\quad x } \)
\(=\frac { cos(x+a-x) }{ cos^{ 2 }x } \)
[\(\because \) cos (A - B) = cos A cos B + sin A sin B]
\(=\frac { cos\quad a }{ cos^{ 2 }x } \).