\(\lim_ { x\rightarrow 0 }{ lim } \frac { sin2x+3x }{ 2x+tan3x } =\lim_ { x\rightarrow 0 }{ lim } \frac { \left( \frac { sin2x }{ 2x } +\frac { 3x }{ 2x } \right) 2x }{ \left( \frac { 2x }{ 3x } +\frac { tan3x }{ 3x } \right) 3x } \)
\(=\left( \lim_ { x\rightarrow 0 }{ lim } \frac { \frac { sin2x }{ 2x } +\frac { 3x }{ 2x } }{ \frac { 2 }{ 3 } +\frac { tan3x }{ 3x } } \right) \)
1
 

cos x + sin x.

\(f\prime (x)=\lim_ { h\rightarrow 0 }{ lim } \frac { log(x+h)^{ 2 }-logx^{ 2 } }{ h } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 2log(x+h)-2logx }{ h } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 2log\left[ 1+\frac { h }{ x } \right] }{ \frac { h }{ x } } x\frac { 1 }{ x } \)
\(=\frac { 2 }{ x } \)

\({ f }^{ ' }(x)=\lim_ { h\rightarrow 0 }{ lim } \frac { sin(x+h+1)-sin(x+1) }{ h } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 2cos\left( \frac { x+h+1+x+1 }{ 2 } \right) sin\left( \frac { x+h+1-x-1 }{ 2 } \right) }{ h } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 2cos\frac { 2x+h+2 }{ 2 } sin\frac { h }{ 2 } }{ 2\times \frac { h }{ 2 } } \)
\(cos(x+1)\)

\(Given \ limit=\lim_ { x\rightarrow 0 }{ lim } \frac { 2sinx-sin2x\quad cosx }{ x^{ 3 } } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { 2sinx-2sinx\quad cosx }{ x^{ 3 } } =\lim_ { x\rightarrow 0 }{ 2lim } { \frac { sinx }{ x } }.\underset { x\rightarrow 0 }{ lim } \left( \frac { 1-cos\quad x }{ { x }^{ 2 } } \right) \)
\(\lim_ { x\rightarrow 0 }{ 2.1lim } \frac { 1-cos\quad x }{ { x }^{ 2 } } =\lim_ { x\rightarrow 0 }{ 2lim } \frac { 2{ sin }^{ 2 }\frac { x }{ 2 } }{ 4\times \frac { { x }^{ 2 } }{ 4 } } \)
1

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

\(\overset{Lt}{x\rightarrow 0}\frac{log(3+x)-log(3-x)}{x}\)=\(\overset{Lt}{x\rightarrow 0}\frac{log[3(1+\frac{x}{3})]-log[3(1-\frac{x}{3})]}{x}\)
⇒ \(\overset{Lt}{x\rightarrow 0}\frac{log3+log(1+\frac{x}{3}-log 3-log(1-\frac{x}{3})}{x} \Rightarrow \overset{Lt}{x\rightarrow 0}\frac{log(1+\frac{x}{3})}{x}-\frac{log(1-\frac{x}{3})}{x}\)
⇒ \(\overset{Lt}{x\rightarrow 0} \frac{log(1+\frac{x}{3})}{3\times \frac{x}{3}}-\frac{log(1-\frac{x}{3})}{\frac{-x}{3}\times -3} \Rightarrow \overset{Lt}{\frac{-x}{3}\times -3} \Rightarrow \overset{Lt}{\frac{x}{3}\rightarrow 0} \frac{1}{3}[\frac{log(1+\frac{x}{3})}{\frac{x}{3}}]+\overset{Lt}{-\frac{x}{3}\rightarrow 0}\frac{1}{3}[\frac{log(1-\frac{x}{3})}{-\frac{x}{3}}]\)
\(\Rightarrow \frac{1}{3}\times 1+\frac{1}{3}\times 1 \Rightarrow\frac{1}{3}+\frac{1}{3}\Rightarrow \frac{2}{3}\)
Hence \(\overset{Lt}{x\rightarrow 0}\frac{log(3+x)-log(3-x)}{x}=\frac{2}{3}\)

Clearly, \(f^{\prime}(-1)=\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{h}\)
\(\left[\because f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\right]\)
\(=\lim _{h \rightarrow 0} \frac{\left[2(-1+h)^{2}+3(-1+h)-5\right]-\left[2(-1)^{2}+3(-1)-5\right]}{h} \)
\(=\lim _{h \rightarrow 0} \frac{\left.\left[2\left(1+h^{2}-2 h\right)-3+3 h-5\right]-[2-3-5\}\right]}{h} \)
\(=\lim _{h \rightarrow 0} \frac{2 h^{2}-h}{h}=\lim _{h \rightarrow 0}(2 h-1)=2(0)-1=-1 \)
\(\text { and } f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}\)
\(\left[\because f^{\prime}(a)=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\right]\)
\(=\lim _{h \rightarrow 0} \frac{\left[2(0+h)^{2}+3(0+h)-5\right]-\left[2(0)^{2}+3(0)-5\right]}{h} \)
\(=\lim _{h \rightarrow 0} \frac{2 h^{2}+3 h}{h}=\lim _{h \rightarrow 0}(2 h+3)=2(0)+3=3 \)
\(\text { Now, } f^{\prime}(0)+3 f^{\prime}(-1)=3-3=0\)
Hence proved.

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

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