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

Given, \(\lim _{ x\rightarrow 1 }{ \frac { f(x)-2 }{ { x }^{ 2 }-1 } =\pi \Rightarrow } \frac { \lim _{ x\rightarrow 1 }{ [f(x)-2] } }{ \lim _{ x\rightarrow 1 }{ { (x }^{ 2 }-1) } } =\pi\)
\( \Rightarrow \lim _{ x\rightarrow 1 }{ [f(x)-2] } =\pi \quad \lim _{ x\rightarrow 1 }{ { (x }^{ 2 }-1) } \)
\(\Rightarrow \lim _{ x\rightarrow 1 }{ f(x)-2=\pi } ({ 1 }^{ 2 }-1) \Rightarrow \lim _{ x\rightarrow 1 }{ f(x)-2=\pi \times 0 } \)
\(\Rightarrow \lim _{ x\rightarrow 1 }{ f(x) } -2=0\Rightarrow \lim _{ x\rightarrow 1 }{ f(x)-2=0 }\Rightarrow \lim _{ x\rightarrow 1 }{ f(x)=2 } \)

\(\lim _{x \rightarrow \pi / 6} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1} \)
\(=\lim _{x \rightarrow \pi / 6} \frac{(2 \sin x-1)(\sin x+1)}{(2 \sin x-1)(\sin x-1)} \)
\(=\lim _{x \rightarrow \pi / 6} \frac{\sin x+1}{\sin x-1} \)
\(=\frac{1+\sin \frac{\pi}{6}}{\sin \frac{\pi}{6}-1}=\frac{1+\frac{1}{2}}{\frac{1}{2}-1}=-3\)

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

\(\frac { dy }{ dx } =\frac { 1 }{ \sqrt [ 2 ]{ x } } -\frac { 1 }{ 2x^{ 3/2 } } \)
Ans: 0

\(put\quad y=\frac { \pi }{ 2 } -x,\quad as\quad x\rightarrow \frac { \pi }{ 2 } ,\quad then\quad y\rightarrow 0\)
\(\therefore \lim_ { x\rightarrow \pi /2 }{ lim } (secx-tanx)\)
\(=\lim_{ y\rightarrow 0 }{ lim } \left[ sec\left( \frac { \pi }{ 2 } -y \right) -tan\left( \frac { \pi }{ 2 } -y \right) \right] \)
\(=\lim_ { y\rightarrow 0 }{ lim } \left( cosec\quad y-cot\quad y \right) \)
\(\left[ \therefore s\left( sec\frac { \pi }{ 2 } -\theta \right) =cosec \ \theta \ and\quad tan\left( \frac { \pi }{ 2 } -\theta \right) =\ cot\quad \theta \right] \)
\(=\lim_ { y\rightarrow 0 }{ lim } \left( \frac { 1 }{ sin\quad y } -\frac { cos\quad y }{ sin\quad y } \right) =\lim_ { y\rightarrow 0 }{ lim } \left( \frac { 1-cos\quad y }{ sin\quad y } \right)\)
\( =\lim_ { y\rightarrow 0 }{ lim } =\frac { 2{ sin }^{ 2 }\frac { y }{ 2 } }{ 2sin\frac { y }{ 2 } cos\frac { y }{ 2 } } \quad \left[ \because { sin }^{ 2 }\frac { \theta }{ 2 } =\frac { 1-cos\quad \theta }{ 2 } \ and \ sin\ \theta =2sin\frac { \theta }{ 2 } cos\frac { \theta }{ 2 } \right] \)
\(=\lim_ { \frac { y }{ 2 } \rightarrow 0 }{ lim } tan\frac { y }{ 2 } =0 \quad [as\quad y\rightarrow 0,\quad then\quad \frac { y }{ 2 } \rightarrow 0]\)

Let \(f(x)=sin\quad x\)
By using first principle of derivative, we have
\(f'(x)=\lim_ { h\rightarrow 0 }{ lim } \frac { f(x+h)-f(x) }{ h } =\lim_ { h\rightarrow 0 }{ lim } \frac { sin(x+h)-sinx }{ h } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 2cos\left( \frac { x+h+x }{ 2 } \right) .sin\left( \frac { x+h-x }{ 2 } \right) }{ h }\)
\(\left[ \because sinC-sinD=2cos\left( \frac { C+D }{ 2 } \right) \times sin\left( \frac { C-D }{ 2 } \right) \right] \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 2cos\left( x+\frac { h }{ 2 } \right) .sin\frac { h }{ 2 } }{ h }\)
\(=\lim_ { h\rightarrow 0 }{ lim } \ cos\left( x+\frac { h }{ 2 } \right) .\lim_ { \frac { h }{ 2 } \rightarrow 0 }{ lim } \frac { sin\frac { h }{ 2 } }{ \frac { h }{ 2 } } \)
\([\therefore h\rightarrow 0\Rightarrow \frac { h }{ 2 } \rightarrow 0]\)
\(=\lim_ { h\rightarrow 0 }{ lim } cos\left( x+\frac { h }{ 2 } \right) \times 1\ \left[ \therefore \lim_ { h\rightarrow 0 }{ lim } \frac { sin\theta }{ \theta } =1 \right] \)
\(=cos(x+0)=cos\ x\)
\(\therefore \frac { d }{ dx } (sinx)=cosx\quad [putting\quad h=0]\)
 

Let f(x) = sec x
By using first principle of derivative, we have
\(f'(x)=\underset { h\rightarrow 0 }{ lim } \frac { f(x+h)-fx() }{ h } \)
\(\therefore f'(x)=\underset { h\rightarrow 0 }{ lim } \frac { sec(x+h)-sec\quad x }{ h }\)
\( =\underset { h\rightarrow 0 }{ lim } \frac { \frac { 1 }{ cos(x+h) } -\frac { 1 }{ cosx } }{ h } \)
\(=\underset { h\rightarrow 0 }{ lim } \frac { cos\quad x-cos(x+h) }{ h\times cos\quad x.cos(x+h) } \)
\(=\underset { h\rightarrow 0 }{ lim } \left[ \frac { -2sin\left( \frac { x+x+h }{ 2 } \right) .sin\frac { (x-x-h) }{ 2 } }{ h.cosx.cos(x+h) } \right] \)
\(\left[ \because cos\quad C-cos\quad D=-2sin\left( \frac { C+D }{ 2 } \right) sin\left( \frac { C-D }{ 2 } \right) \right] \)
\(=\underset { h\rightarrow 0 }{ lim } \left[ \frac { -2sin\left( x+\frac { h }{ 2 } \right) .\left( -sin\frac { h }{ 2 } \right) }{ h.cosx\quad cos(x+h) } \right] \)
\(=\underset { h\rightarrow 0 }{ lim } \frac { sin\left( x+\frac { h }{ 2 } \right) }{ cos(x+h).cos\quad x } .\underset { h\rightarrow 0 }{ lim } \frac { sin\frac { h }{ 2 } }{ \frac { h }{ 2 } } \)
\(=\frac { sin\quad x }{ cos^{ 2 }\quad x } \times 1\)
\(=\frac { sin\quad x }{ cos\quad x } .\frac { 1 }{ cos\quad x } =tanx.sec\quad x\)

Let f(x) = tan x
Then, by first principle of derivative, we get
\(f'(x)=\lim_ { h\rightarrow 0 }{ lim } \frac { f(x+h)-f(x) }{ h } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \quad \frac { tan(x+h)-tanx }{ h } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 1 }{ h } \left[ \frac { sin(x+h) }{ cos(x+h) } -\frac { sin\quad x }{ cos\quad x } \right] \)
\(=\lim_{ h\rightarrow 0 }{ lim } \frac { 1 }{ h } \left[ \frac { sin(x+h-x) }{ cos(x+h)cos\quad x } \right] \)
\([\because sin\quad AcosB-cos\quad A\quad sin\quad B=sin(A-B)]\)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { sin\quad h }{ h } .\lim_ { h\rightarrow 0 }{ lim } \frac { 1 }{ cos(x+h)cos\quad x } \)
\(=1.\frac { 1 }{ cos(x+0)cos\quad x } \)
\(=\frac { 1 }{ cos^{ 2 }x } =sec^{ 2 }x\quad [\because \lim_{ h\rightarrow 0 }{ lim } \frac { sin\quad \theta }{ \theta } =1]\)
\(hence,\quad f'(x)or\quad \frac { d }{ dx } (tanx)=sec^{ 2 }x\)

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