\(\lim_{x\rightarrow 0} \frac{log_{e} (1+2x)}{x}\)
⇒ \(\lim_{2x\rightarrow 0} \frac{2log_{e}(1+2x)}{2x}​=2\times 1​\)
\(\lim_{t\rightarrow 0} \frac{log_{e}(1+t)}{t}=1​​]=2\)

\(\lim_{x\rightarrow 0} \frac{x(e^{x}-1)}{1-cos x}\)
⇒ \(\lim_{x\rightarrow 0} [\frac{\frac{x(e^{x-1})}{x^{2}}}{\frac{1-cosx}{x^{2}}}]\)
⇒ \(\lim_{x\rightarrow 0} [\frac{\frac{e^{x}-1}{x}}{\frac{2sin^{2}x/2}{x^{2}}}]\)
⇒ \(\frac{{Lim}_{x\rightarrow 0}[\frac{e^{x}-1}{x}]}{{Lim}_{x\rightarrow 0}[\frac{2sin^{2}x/2}{4\times \frac{x^{2}}4{}}]}\)
⇒ \(\lim_{\frac{x}{2}\rightarrow 0}\frac{1}{\frac{1}{2}[\frac{sinx/2}{\frac{x}{2}}]^{2}} \Rightarrow \frac{1}{\frac{1}{2}\times (1)^{2}} \Rightarrow 2\).

\(\underset { x\rightarrow o }{ Lim } \frac { { e }^{ { x }^{ 2 } }-cosx }{ { x }^{ 2 } } \)
\(=\underset { x\rightarrow o }{ Lim } \frac { { e }^{ { x }^{ 2 } }-1-cosx+1 }{ { x }^{ 2 } } \)
\(=\underset { x\rightarrow o }{ Lim } \frac { { e }^{ { x }^{ 2 } }-1 }{ { x }^{ 2 } } +\frac { 1-cosx }{ { x }^{ 2 } } \)
\(=\underset { x^{ 2 }\rightarrow o }{ Lim } \left[ \frac { { e }^{ { x }^{ 2 } }-1 }{ { x }^{ 2 } } \right] +\underset { x\rightarrow o }{ Lim } \left[ \frac { { 2sin }^{ 2 }x/2 }{ \frac { { x }^{ 2 } }{ 4 } \times 4 } \right] \)
\(\Rightarrow 1+\frac { 1 }{ 2 } \underset { \frac { x }{ 2 } \rightarrow 0 }{ Lim } \left[ \frac { sin\frac { x }{ 2 } }{ \frac { x }{ 2 } } \right] ^{ 2 }\Rightarrow 1+\frac { 1 }{ 2 } \times 1\)
\(\Rightarrow 1+\frac { 1 }{ 2 } \Rightarrow \frac { 3 }{ 2 } \)

\(\lim_{ x\rightarrow o }\frac { { e }^{ tanx }-1 }{ log(1+x) } \)
\(=\lim_ { x\rightarrow o } \frac { \frac { { e }^{ tanx }-1 }{ tanx } \times tanx }{ \frac { log(1+x) }{ x } \times x } \)
\(\Rightarrow \lim_ { x\rightarrow o } \left( \frac { { e }^{ tanx }-1 }{ tanx } \right) \)
\(\therefore tanx\rightarrow 0\)
\(\lim_{ x\rightarrow o } \frac { log(1+x) }{ x } \times \lim_ { x\rightarrow o } \left[ \frac { tanx }{ x } \right] \)
\(\Rightarrow \frac { 1 }{ 1 } \times 1\Rightarrow 1\) 

\(\lim _{x \rightarrow 0}\left\{\left(\frac{a^{x}-1}{\sin x}\right)-\left(\frac{b^{x}-1}{\sin x}\right)\right\} \)
\(= \lim _{x \rightarrow 0}\left(\frac{a^{x}-1}{x} \times \frac{x}{\sin x}\right)-\lim _{x \rightarrow 0}\left(\frac{b^{x}-1}{x} \times \frac{x}{\sin x}\right) \)
\(= \lim _{x \rightarrow 0} \frac{a^{x}-1}{x} \times \lim _{x \rightarrow 0} \frac{x}{\sin x}-\lim _{x \rightarrow 0} \frac{b^{x}-1}{x} \times \lim _{x \rightarrow 0} \frac{x}{\sin x} \\\)
\(A n s \cdot \log \left(\frac{a}{b}\right)\)

\(\lim _{x \rightarrow 0} \frac{\log \left\{5\left(1+\frac{x}{5}\right)\right\}-\log \left\{5\left(1-\frac{x}{5}\right)\right\}}{x}\)
\(=\lim _{x \rightarrow 0} \frac{\left\{\log 5+\log \left(1+\frac{x}{5}\right)\right\}-\left\{\log 5+\log \left(1-\frac{x}{5}\right)\right\}}{x}\)
\(=\lim _{x \rightarrow 0} \frac{1}{5} \frac{\log \left(1+\frac{x}{5}\right)}{\frac{x}{5}}-\lim _{x \rightarrow 0} \frac{\log \left(1-\frac{x}{5}\right)}{-\frac{x}{5}} \cdot \frac{1}{(-5)}\)
\(\text { Ans. } \frac{2}{5}\)

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

on putting  x=2 we get the form \( \frac {0}{0}\). So, let us first factorise it .
Consider, \( \lim _{x \rightarrow 2} \frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^{2}} \)
\(=\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)}=\frac{2+2}{2(2-2)}=\frac{4}{0}\)
which is not defined.
\(\therefore \lim _{x \rightarrow 2}\left[\frac{x^{2}-4}{x^{3}-4 x^{2}+4 x}\right]\) dose not exist

\(\lim _{x \rightarrow 2} \frac{x^{2}-x \log x+2 \log x-4}{x-2} \)
\(=\lim _{x \rightarrow 2} \frac{\left(x^{2}-4\right)-\log x(x-2)}{x-2} \)
\(=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)-\log x(x-2)}{(x-2)} \)
\(=\lim _{x \rightarrow 2} \frac{(x-2)[x+2-\log x]}{(x-2)} \)
\(=\lim _{x \rightarrow 2}[x+2-\log x]=2+2-\log 2 \)
\(=4-\log 2\)

we have, 7t⁴ - 2t³ - 8t - 5
On differentiating both sides w.r.t. t, we get 
\(\frac { du }{ dt } =\frac { d }{ dt } \left[ { 7t }^{ 4 }-{ 2t }^{ 3 }-8t-5 \right] \)
\(=7\left( { 4t }^{ 3 } \right) -2\left( { 3t }^{ 2 } \right) -8(1)-0\)  \([\because \frac { d }{ dx } ({ x }^{ n })={ nx }^{ n-1 }]\)
= 28t³ - 6t² - 8
Now, \(\left( \frac { du }{ dt } \right) _{ t=2 }=28(2)^{ 3 }-6(2)^{ 2 }-8\)

\(= 224-24-8=192\)

On putting x =\( \frac{1} {2}\), we get the form\( \frac{0}{0}\)
So,let us first factorise it
Consider, \(\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x+1)(2 x-1)}{(2 x-1)}\) [using factorisation method]
\(=\lim _{x \rightarrow \frac{1}{2}}(2 x+1) \)
\(=2\left(\frac{1}{2}\right)+1=2\)

Let f(x) = a^x.
By using first principle of derivative, we have
\(\therefore f'(x)=\lim_ { h\rightarrow 0 }{ lim } \frac { f(x+h)-f(x) }{ h } \)
\(\Rightarrow f'(x)=\lim_ { h\rightarrow 0 }{ lim } \frac { a^{ x+h }-a^{ x } }{ h } =\lim_{ h\rightarrow 0 }{ lim } \frac { { a }^{ x }{ a }^{ h }-{ a }^{ x } }{ h } \)
\(\Rightarrow f'(x)={ a }^{ x }\lim_ { h\rightarrow 0 }{ lim } \left( \frac { { a }^{ h }-1 }{ h } \right) ={ a }^{ x }log_{ e }a\quad \left[ \lim_ { x\rightarrow 0 }{ \because lim } \frac { { a }^{ x }-1 }{ x } =log_{ e }a \right] \)

\(\lim_ { x\rightarrow \pi /2 }{ lim } \frac { 1+cos 2x }{ { \left( \pi -2x \right) }^{ 2 } } =\lim_ { h\rightarrow 0 }{ lim } \frac { 1+cos\ 2\left( \frac { \pi }{ 2 } +h \right) }{ { \left[ \pi -2\left( \frac { \pi }{ 2 } +h \right) \right] }^{ 2 } } \)
\(\left[ putting\quad x=\frac { \pi }{ 2 } +h,\quad as\quad x\quad \rightarrow \frac { \pi }{ 2 } ,\quad then\quad h\rightarrow 0 \right] \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 1+cos\quad (\pi +2h) }{ { \left( \pi -\pi -2h \right) }^{ 2 } } \)
\(=\lim_ { h\rightarrow 0 }{ lim } \frac { 1-cos\quad 2h }{ { 4h }^{ 2 } } \quad \left[ \frac { 0 }{ 0 } form \right] \)
\(=\lim_{ x\rightarrow 0 }{ lim } \frac { 1-\left( 1-2{ sin }^{ 2 }h \right) }{ 4{ h }^{ 2 } } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { 2{ sin }^{ 2 }h }{ { 4h }^{ 2 } } \quad \left[ \therefore \ cos \ 2 \theta =1-2{ sin }^{ 2 }\theta \right] \)
\(=\frac { 2 }{ 4 } \lim_ { h\rightarrow 0 }{ lim } { \left( \frac { sin\quad h }{ h } \right) }^{ 2 }=\frac { 2 }{ 4 } \times 1=\frac { 1 }{ 2 } \quad \left[ \because \quad \lim_ { \theta \rightarrow 0 }{ lim } \frac { sin\quad \theta }{ \theta } =1 \right] \)

\(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}=\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x} \times \frac{3}{3}\)
[ multiplying numerator and denominator by 3]
\(=3 \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{3 x}\) ..(i)
Let h = 3x. Then \(x\rightarrow 0 \Rightarrow h\rightarrow 0\)
Now, from E.q (i), we get
\(\left.\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}=3 \lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=3(1) \quad[ \because \lim _{\theta \rightarrow 0} \frac{e^{\theta}-1}{\theta}=1\right]\)
= 3

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