\(\overset{Lt}{x\rightarrow 0} [tan(\frac{\pi}{4}+x)]^{\frac{1}{cotx}}\)=\(\overset{Lt}{x\rightarrow 0} [\frac{tan\frac{\pi}{4}+tanx}{1-tan\frac{\pi}{4}.tanx}]^{cotx}\)
=\(\overset{Lt}{x\rightarrow 0}[\frac{1+tanx}{1-tanx}]^{\frac{1}{tanx}}\) ⇒ \(\frac { \underset { \underset { tanx\rightarrow 0 }{ x\rightarrow 0 } }{ Lt } { \left[ [1+tanx \right] }^{ \frac { 1 }{ tanx } } }{ \underset { \underset { -tanx\rightarrow 0 }{ x\rightarrow 0 } }{ Lt } { \left[ { \left( 1-tanx \right) }^{ -\frac { 1 }{ taanx } } \right] }^{ -1 } } =\frac { e }{ { e }^{ -1 } } ={ e }^{ 2 }\)
Hence \(\overset{Lt}{x\rightarrow 0} [tan(\frac{\pi}{4}+x)]^{cotx}=e^{2}.\)

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

Let f(x) = cot\(\sqrt{x}\)
f(x+h)=cot\(\sqrt{x+h}\)
\(\therefore \frac{d}{dx} f(x)=\overset{lim}{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)
=\(\overset{lim}{h\rightarrow 0}\frac{cot\sqrt{x+h}-cot \sqrt{x}}{h}\)
=\(\overset{lim}{h\rightarrow 0} \frac{-sin (\sqrt{x+h}-\sqrt{x})}{(x+h-x)sin\sqrt{x+h}.sin\sqrt{x}}\)
=\(\overset{lim}{h\rightarrow 0}\frac{-sin (\sqrt{x+h}-\sqrt{x})}{(\sqrt{x-h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}\)
=\(\overset{lim}{h\rightarrow 0}\frac{sin(\sqrt{x+h}-\sqrt{x})}{(\sqrt{x+h}-\sqrt{x})}\)
\(\overset{lim}{h\rightarrow 0}\frac{-1}{(\sqrt{x+h}+\sqrt{x})sin\sqrt{x+h}.sin\sqrt{x}}\)
=\(\frac{-1}{2\sqrt{x}sin\sqrt{x}sin\sqrt{x}}=\frac{-cosec^{2}\sqrt{x}}{2\sqrt{x}}\)

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

\(\underset { x\rightarrow o }{ Lim } \frac { { 27 }^{ x }-{ 9 }^{ x }-{ 3 }^{ x }+1 }{ 1-cosx } \)
\(=\underset { x\rightarrow o }{ Lim } \frac { ({ 9 }^{ x }-1)({ 3 }^{ x }-1) }{ 1-cosx } \)
\(\Rightarrow \underset { x\rightarrow o }{ Lim } \frac { \frac { ({ 9 }^{ x }-1)({ 3 }^{ x }-1) }{ { x }^{ 2 } } }{ \frac { 1-cosx }{ { x }^{ 2 } } } \)
\(\Rightarrow \underset { x\rightarrow o }{ Lim } \frac { \left( \frac { { 9 }^{ x }-1 }{ x } \right) \left( \frac { { 3 }^{ x }-1 }{ x } \right) }{ \left[ 2\left( \frac { sinx }{ x } \right) ^{ 2 } \right] } \)
\(\Rightarrow \frac { { log }_{ e }9\times { log }_{ e }3 }{ 2 } \Rightarrow \frac { { log }_{ e }{ 3 }^{ 2 }\times { log }_{ e }{ 3 } }{ 2 } \)
\(\Rightarrow \frac { { 2({ log }_{ e }3) }^{ 2 } }{ 2 } \Rightarrow \left( { log }_{ e }3 \right) ^{ 2 }\)

here f(x)=\((px+q)\left( \frac { r }{ x } +s \right) \)
\(\therefore f(x)=(px+q)\left( \frac { r }{ x } +s \right) \)
\( =(px+q)\frac { d }{ dx } \left( \frac { r }{ x } +s \right) +\left( \frac { r }{ x } +s \right) \frac { d }{ dx } (px+q)\)
\(=(px+q)\left( \frac { -r }{ { x }^{ 2 } } \right) +\left( \frac { r }{ x } +s \right) (p)\)
\(=\frac { -pr }{ x } \frac { -qr }{ { x }^{ 2 } } +\frac { pr }{ x } +ps\)
\(=\frac { -qr }{ { x }^{ 2 } } +ps\)

\(\text { Let } f(x)=x+a \text { . Accordingly, } f(x+h)=x+h+a\)
\(\text { By first principle, }\)
\(f^{\prime}(x) =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \)
\(=\lim _{h \rightarrow 0} \frac{x+h+a-x-a}{h} \)
\(=\lim _{h \rightarrow 0}\left(\frac{h}{h}\right) \)
\(=\lim _{h \rightarrow 0}(1) \)
\(=1\)

(i) Here f(x) = - x Then if (x + h) = - (x + h) We known that
\(f\left( x \right) =\lim _{ h\rightarrow o }{ \frac { f(x+h)-f(x) }{ h } } \)
\(\Rightarrow f(x)=\lim _{ h\rightarrow o }{ \frac { -(x+h)-(-x) }{ h } } \)
\(=\lim _{ h\rightarrow o }{ \frac { -h }{ h } } =-1\)
Here f(x)=(-x)^-1=-\(-\frac { 1 }{ x } \)
f(x+h)=-\(\frac { 1 }{ x+h } \)
We know that
\(f(x)=\lim _{ h\rightarrow o }{ \frac { f(x+h)-f(x) }{ h } } \)
\(\Rightarrow f\left( x \right) =\lim _{ h\rightarrow o }{ \frac { -\frac { 1 }{ x+h } -\left( -\frac { 1 }{ x } \right) }{ h } } \)
\( =\lim _{ h\rightarrow o }{ \frac { -x+x+h }{ hx(x+h) } } \)
\(=\lim _{ h\rightarrow o }{ \frac { h }{ hx(x+h) } } =\frac { 1 }{ { x }^{ 2 } } \)
(iii) Here f(x)=sin(x+1)
Then f(x+h)=sin(x+h+1)
(iii) We know that
\(f(x)\lim _{ h\rightarrow o }{ \frac { f(x+h)-f(x) }{ h } } \)
\(\Rightarrow f(x)=\lim _{ h\rightarrow o }{ \frac { sin(x+h+1)-sin(x+1) }{ h } } \)
\(=\lim _{ h\rightarrow o }{ \frac { 2cos\left( \frac { 2x+h+2 }{ 2 } \right) sin\frac { h }{ 2 } }{ h } } \)
\(=\lim _{ h\rightarrow o }{ \frac { cos\left( x+1+\frac { h }{ 2 } \right) sin\left( \frac { h }{ 2 } \right) }{ \frac { h }{ 2 } } } \)
\( =cos(x+1)\)
(iv) Here \(f\left( x \right) =cos\left( x-\frac { \pi }{ 8 } \right) \)
Then \(f(x+h)=cos\left( x+h-\frac { \pi }{ 8 } \right) \)
We know that
\(f\left( x \right) =\lim _{ h\rightarrow o }{ \frac { f(x+h)-f(x) }{ h } } \)
\(\Rightarrow f(x)=\lim _{ h\rightarrow o }{ \frac { cos\left( x+h-\frac { \pi }{ 8 } \right) -cos\left( x-\frac { \pi }{ 8 } \right) }{ h } } \)
\(=\lim _{ h\rightarrow 0 }{ \frac { -2sin\left( x-\frac { \pi }{ 8 } +\frac { h }{ 2 } \right) sin\left( \frac { h }{ 2 } \right) }{ \frac { h }{ 2 } } }\)
\(=\lim _{ h\rightarrow o }{ \frac { -sin\left( x-\frac { \pi }{ 8 } +\frac { h }{ 2 } \right) sin\left( \frac { h }{ 2 } \right) }{ \frac { h }{ 2 } } }\)
\(=-sin\left( x-\frac { \pi }{ 8 } \right) \)

⇒ \(f(x)=\left\{\begin{array}{ll} |x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0 \end{array}\right.\)
\(\text { When } a<0 \text { , } \)
\(\lim _{x \rightarrow \sigma^{-}} f(x) =\lim _{x \rightarrow a^{-}}(|x|+1) \)
\(=-a+1 \)
\(\lim _{x \rightarrow c^{+}} f(x) =\lim _{x \rightarrow a^{+}}(|x|+1) \)
\(=\lim _{x \rightarrow a}(-x+1) \)
\(=-a+1\)
\(\therefore \lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a^{+}} f(x)=-a+1\)
\(\text { Thus, limit of } f(x) \text { exists at } x=a, \text { where } a<0 \text { . }\)
\(\text { When } a>0\)|
\(\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{-}}(|x|-1)\)
\(=\lim _{x \rightarrow a}(x-1) \quad[0
\(=a-1 \)
\(\lim _{x \rightarrow a^{+}} f(x) =\lim _{x \rightarrow a^{+}}(|x|-1) \)
\(=\lim _{x \rightarrow a}(x-1) [0
\(=a-1\)
\(\therefore \lim _{x \rightarrow a^{+}} f(x)=\lim _{x \rightarrow d^{+}} f(x)=a-1\)
\(\text { Thus, limit of } f(x) \text { exists at } x=a, \text { where } a>0 .\)
\(\text { Thus, } \lim _{x \rightarrow a} f(x) \text { exists for all } a \neq 0 \text { . }\)
\(f(x)=\left\{\begin{array}{ll} |x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0 \end{array}\right.\)
\(\text { When } a<0 \text { , } \)
\(\lim _{x \rightarrow 0^{-}} f(x) =\lim _{x \rightarrow 0^{-}}(|x|+1) \)
\(=\lim _{x \rightarrow 0}(-x+1) \quad[\text { If } x<0,|x|=-x] \)
\(=-0+1 \)
\(=1 \)
\(\lim _{x \rightarrow 0^{+}} f(x) =\lim _{x \rightarrow 0^{+}}(|x|-1) \)
\(=\lim _{x \rightarrow 0}(x-1) \quad[\text { If } x>0,|x|=x] \)
\(=0-1 \)
\(=-1\)
\(\text { Here, it is observed that } \lim _{x \rightarrow 0} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x) \text { . }\)
\(\therefore \lim _{x \rightarrow 0} f(x) \text { does not exist. }\)

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

\(\text { Let } f(x)=\frac{1}{a x^{2}+b x+c}\)
\(\text { By quotient rule, }\)
\(f^{\prime}(x) =\frac{\left(a x^{2}+b x+c\right) \frac{d}{d x}(1)-\frac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}} \)
\(=\frac{\left(a x^{2}+b x+c\right)(0)-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}} \)
\(=\frac{-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}\)

Here f(x)=\(\frac { ax+b }{ cx+d } \)
\(\therefore f(x)=\frac { d }{ dx } \left[ \frac { ax+b }{ cx+d } \right] \)
\(=\frac { (cx+d)\frac { d }{ dx } (ax+b)-(ax+b)\frac { d }{ dx } (cx+d) }{ { (cx+d) }^{ 2 } } \)
\(=\frac { (cx+d)(a)-(ax+b)(c) }{ { (cx+d) }^{ 2 } } \)
\(=\frac { acx+ad-acx-bc }{ { (cx+d) }^{ 2 } } \)
\(=\frac { ad-bc }{ { (cx+d) }^{ 2 } } \)