\(LHL=\lim _{ X\rightarrow { -1 }^{ - } }{ (x+2)= } \lim _{ h\rightarrow 0 }{ (-1-h+2)=1 } \)
\(RHL=\lim _{ x\rightarrow { 1 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { -1 }^{ + } }{ { cx }^{ 2 } } =\lim _{ h\rightarrow 0 }{ c{ (-1+h })^{ 2 } } \)
Ans. c = 1

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

\(\lim _{x \rightarrow \pi / 6} \frac{\sqrt{3} \sin x-\cos x}{x-\frac{\pi}{6}}\)
\(=\lim _{x \rightarrow \pi / 6} \frac{2\left(\sin x \cos \frac{\pi}{6}-\cos x \sin \frac{\pi}{6}\right)}{\left(x-\frac{\pi}{6}\right)}\)
\(=2 \lim _{x \rightarrow \pi / 6} \frac{\sin \left(x-\frac{\pi}{6}\right)}{\left(x-\frac{\pi}{6}\right)} \)
\([\because \sin A \cos B-\cos A \sin B=\sin (A-B)] \)
\(=2 \left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1 \text { and } x \rightarrow \frac{\pi}{6} \Rightarrow\left(x-\frac{\pi}{6}\right) \rightarrow 0\right]\)

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

\(Let\quad f(x)=\frac { sinx+cosx }{ sinx-cosx } \)
On differentiating both sides w.r.t. x,we get
\({ f }^{ ' }(x)=\frac { \left[ (sinx-cosx)\frac { d }{ dx } (sinx+cosx)\\ -(sinx+cosx)\quad \frac { d }{ dx } (sinx-cosx) \right] }{ \left( sinx-cosx \right) ^{ 2 } } \)
\(\left[ \because \frac { d }{ dx } (\frac { f(x) }{ g(x) } )=\frac { g(x)\frac { d }{ dx } f(x)-f(x)\frac { d }{ dx } g(x) }{ \left[ g(x) \right] ^{ 2 } } \right]\)
\(=\frac { \left[ \left( sinx-cosx \right) \left[ \frac { d }{ dx } (sinx)+\frac { d }{ dx } (cosx) \right] \\ -(sinx+cosx)\left[ \frac { d }{ dx } (sinx)-\frac { d }{ dx } (cosx) \right] \right] }{ ({ sinx-cosx) }^{ 2 } } \)
\(=\frac { (sinx-cosx)(cosx-sinx)-(sinx+cosx)(cosx+sinx) }{ ({ sinx-cosx) }^{ 2 } } \)
\(\left[ \because \frac { d }{ dx } (sinx)=cosx\quad and\frac { d }{ dx } (cosx)=-sinx \right] \)
\(=\frac { -({ sinx-cosx) }^{ 2 }-({ sinx-cosx) }^{ 2 } }{ ({ sinx-cosx) }^{ 2 } } \)
\(=-\left[ \frac { ({ sinx }-cosx)^{ 2 }+({ sinx+cosx) }^{ 2 } }{ ({ sinx-cosx) }^{ 2 } } \right] \)
\( =\frac { \left[ -({ sin }^{ 2 }x+{ cos }^{ 2 }x-2sinxcosx+{ sin }^{ 2 }x\\ +{ cos }^{ 2 }x+2sinxcosx) \right] }{ ({ sinx-cosx })^{ 2 } } \)
\(=\frac { -2({ sin }^{ 2 }x+{ cos }^{ 2 }x) }{ (sinx-cosx)^{ 2 } } =\frac { -2 }{ ({ sinx-cosx) }^{ 2 } } \)
\(\left[ \because { sin }^{ 2 }\theta +{ cos }^{ 2 }\theta =1 \right] \)

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

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

Let \(y=\frac { secx+tanx }{ secx-tanx } =\frac { \frac { 1 }{ cosx } +\frac { sinx }{ cosx } }{ \frac { 1 }{ cosx } -\frac { sinx }{ cosx } } \Rightarrow y=\frac { 1+sinx }{ 1-sinx }\) 
On differentiating both sides w.r.t. x,we get 
\(\frac { dy }{ dx } =\frac { d }{ dx } \left( \frac { secx+tanx }{ secx-tanx } \right) =\frac { d }{ dx } \left( \frac { 1+sinx }{ 1-sinx } \right)\) 
\(​​​​​​​=\frac { \left( 1-sinx \right) \frac { d }{ dx } \left( 1+sinx \right) -\left( 1+sinx \right) \frac { d }{ dx } \left( 1-sinx \right) }{ ({ 1-sinx) }^{ 2 } } \)
\(=\frac { \left( 1-sinx \right) \left( 0+cosx \right) -\left( 1+sinx \right) \frac { d }{ dx } \left( 1-sinx \right) }{ (1-sinx)^{ 2 } } \)  [using quotient rule of derivative] 
\(=\frac { \left( 1-sinx \right) \left( 0+cosx \right) -\left( 1+sinx \right) \left( 0-cosx \right) }{ ({ 1-sinx })^{ 2 } } \)
\(=\frac { 2cosx }{ (1-sinx)^{ 2 } } \)

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