$\lim _{x \rightarrow 0} \frac{e^{5 x}+e^{-5 x}-2}{x^{2}}$
$=\lim _{x \rightarrow 0} \frac{5 e^{5 x}-5 e^{-5 x}}{2 x}$ [L'Hopital's rule]
$=\lim _{x \rightarrow 0} \frac{25 e^{5 x}+25 e^{-5 x}}{2}$ [L'Hopital's rule]
$=\frac{25+25}{2}=25$

$\lim _{h \rightarrow 0} \frac{\frac{1}{2+h}+0}{1}=\frac{1}{2}$

$\lim _{h \rightarrow 0} \frac{\ln (2+h)-\ln 2}{h}=\lim _{h \rightarrow 0} \frac{\ln \left(\frac{2+h}{2}\right)}{h}$
$=\lim _{h \rightarrow 0} \frac{\ln \left(1+\frac{h}{2}\right)}{h}=\frac{1}{2} \lim _{h \rightarrow 0} \frac{\ln \left(1+\frac{h}{2}\right)}{\frac{h}{2}}$
$=\frac{1}{2} \cdot 1=\frac{1}{2}$

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\left(\frac{\pi}{2}-x\right)^{2}}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-2\left(\frac{\pi}{2}-x\right)}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\sin x}{2(0-1)}=\frac{1}{2}=0.5$

$\lim _{x \rightarrow-4} \frac{x+4}{x^{2}+2 x-8}=\lim _{x \rightarrow-4} \frac{1}{2 x+2}$[L'Hopital's Rule]
$=\frac{1}{2(-4)+2}=-\frac{1}{6}$

$\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin x}=\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}-2}{1-\cos x}$
$=\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin x}=\lim _{x \rightarrow 0} \frac{e^{x}+e^{-x}}{\cos x}$
$=\frac{e^{0}+e^{-0}}{\cos 0}=\frac{1+1}{1}=2$ [L'Hopital's Rule]

$\lim_{x\rightarrow\frac{\pi}{2}}\frac{1-\sin x}{\cos x}$ limit বসালে 0 আসে, L'Hopital's Rule প্রয়োগ করে $\lim _{x \rightarrow \frac{\pi}{2}} \frac{0-\cos x}{-\sin x}=0$

$\lim _{x \rightarrow 0} \frac{x}{\sin x} \cdot \lim _{x \rightarrow 0}(\cos x+\cos 2 x)$
$=1 \times(\cos 0+\cos 0)=1+1=2$

$\lim _{x \rightarrow 0} \frac{\sqrt{1+2 x}-\sqrt{1-3 x}}{x}\left[\frac{0}{0}\right]$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+2 x}} \times 2-\frac{1}{2 \sqrt{1-3 x}}(-3)}{1}$
$=\frac{1+\frac{3}{2}}{1}=\frac{5}{2}$

$\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}\left[\frac{0}{0} \text { form }\right]$
$\Rightarrow \lim _{x \rightarrow 0} \frac{1}{1+x}=1$

$\lim _{x \rightarrow 0} \frac{a^{x}-1}{x} ; \frac{0}{0}$ আকার বলে L'Hopital rule প্রয়োগ করে পাই, $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}$
$=\lim _{x \rightarrow 0} \frac{a^{x} \ln a}{1}=\ln a$

$\lim_{x\rightarrow0} \frac{\sin x^{\circ}}{x}$
$=\frac{\pi}{180}\times\lim_{x\rightarrow0}\frac{\sin\frac{\pi\times x}{\pi80}}{\frac{\pi\times x}{180}}$
$=\frac{\pi}{180}$ [রেডিয়ান এককে করবে, $1^{\circ}=\left(\frac{\pi}{180}\right)^{c}]$

$\lim _{x \rightarrow 0} \frac{e^{\cos x}}{\cos x}=\frac{e^{\cos 0}}{\cos 0}=\frac{e^{1}}{1}=e$

$\lim _{x \rightarrow \infty} \frac{x^{2}-4}{2+x-4 x^{2}}=\lim _{x \rightarrow \infty} \frac{1-\frac{4}{x^{2}}}{\frac{2}{x^{2}}+\frac{1}{x}-4}=-\frac{1}{4}$ [ $x^{2}$এর সহগের অনুপাত]

$\lim _{x \rightarrow 0}(1+4 x)^{\frac{3 x+2}{x}}=\lim _{x \rightarrow 0}(1+4 x)^{3+\frac{2}{x}}$
$=e^{4 \cdot 2}=e^{8}$ $\left[\because \lim _{x \rightarrow 0}(1+a x)^{\frac{b}{x}+c}=e^{a b}\right]$

$\lim _{x \rightarrow 0} \frac{\cos 2 x-\cos 3 x}{x^{2}}$
$=\lim _{x \rightarrow 0} \frac{-2 \sin 2 x+3 \sin 3 x}{2 x}$
$=\lim _{x \rightarrow 0} \frac{-4 \cos 2 x+9 \cos 3 x}{2}=\frac{9-4}{2}=\frac{5}{2}$

L — Hopital's law ব্যাবহার করে
$\lim _{x \rightarrow 0} \frac{\sin 7 x-\sin x}{\sin 6 x}$
$=\lim _{x \rightarrow 0} \frac{7 \cos 7 x-\cos x}{6 \cos 6 x}=\frac{7-1}{6}=1$

$y=\frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}}$
$=\frac{\sin x+\cos x}{\sqrt{\sin ^{2} x+\cos ^{2} x+2 \sin x \cdot \cos x}}=1$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0$

$\frac{dy}{dx} = \frac{d}{dx} (cx + cx^2) = c + 2cx$
$\therefore$ মূলবিন্দুর জন্য $\frac{dy}{dx} = c + 2c.0$
$\therefore$ $c = \tan 30^\circ = \frac{1}{\sqrt{3}}$

$y = 4x^2 + 3x - 5$
$\Rightarrow \frac {dy}{dx} = 8x + 3$
$\therefore (\frac {dy}{dx})_{(1,2)} = 8 \times 1 + 3 = 11$

$y = x + \frac{1}{x}$
$\Rightarrow \frac {dy}{dx} = 1 - \frac{1}{x^2} = 0$
$\Rightarrow x^2 = 1$
$\therefore x = \pm 1$

$y = ln(1+x),$
$\frac{dy}{dx} = \frac{1}{1+x} = (1+x)^{-1}$
$\frac{d^2y}{dx^2} = -1(1+x)^{-1-1} = - \frac{1}{(1+x)^2}$

$x^2 + 3xy + 5y^2 = 1$
$\Rightarrow 2x + 3y + 3x\frac{dy}{dx} + 10y\frac{dy}{dx} = 0$
$\Rightarrow \frac{dy}{dx} = - \frac{2x + 3y}{3x + 10y}$

$e^{xy+1} = 5;$
$xy + 1 = ln5$
$\Rightarrow xy = ln5 - 1$
$\therefore x \frac{dy}{dx} + y = 0$
$\Rightarrow \frac{dy}{dx} = - \frac{y}{x}$

$\frac{dy}{dx} = \frac{d}{dx} {sin^{-1} (sinx)} = \frac{d}{dx} (x) = 1$

$y = tan^{-1} \frac{1 + x}{1 - x} = tan^{-1} \frac{1 + x}{1 - 1.x}$
$= tan^{-1} 1 + tan^{-1} x = \frac{\pi}{4} + tan^{-1} x$
$\therefore \frac{dy}{dx} = \frac{1}{1 + x^2}$

$e^{2\log x + 1} = e^{2\log x^2} .e = x^2.e$
$\therefore \frac{d}{dx} (x^2.e) = 2xe$

$\frac{d}{dx} (\log (x - \sqrt{x^2 - 1}))$
$= \frac{1}{x - \sqrt{x^2 - 1}} \times (1 - \frac{1}{2\sqrt{x^2 -1}} \times 2x)$
$= \frac{1}{x - \sqrt{x^2 - 1}} \times \frac{\sqrt{x^2 - 1} - x}{\sqrt{x^2 - 1}} = \frac{-1}{\sqrt{x^2 - 1}}$

$y = \log _{x} a = \log _{e} a \times \log _{x} e = ln a \times \frac {1}{ln x}$
$\therefore \frac{dy}{dx} = ln a \times \frac{d}{dx} (\frac {1}{ln x}) = - \frac{1}{(ln x)^2} \times \frac{1}{x} \times ln a = - \frac{ln a}{x(ln x)^2}$

$\frac{d}{dx} {(\frac{1}{1+ 2x})} = (-1) \frac{2}{(1 + 2x)^2} = \frac{-2}{(1 + 2x)^2}$

$\frac{d}{dx} {\tan^{-1} (e^{x})} = \frac{1}{1 + e^{2x}} \times e^{x}$
$= \frac{e^{x}}{1 + e^{2x}}$

$\frac{d}{dx} (\tan^{-1} \frac{x}{5}) = \frac{1}{1 + (\frac{x}{5})^2} \times \frac{1}{5}$
$= \frac{1 \times 25}{25 + x^2} \times \frac{1}{5} = \frac{5}{25 + x^2}$

$y = \log \sin x^2$
$\Rightarrow \frac{dy}{dx} = \frac{1}{\sin x^2} \times \frac{d}{dx} (\sin x^2)$
$\Rightarrow \frac{dy}{dx} = \frac{1}{\sin x^2} \times \cos x^2 \times \frac{d}{dx} (x^2)$
$\therefore \frac{dy}{dx} = 2x \cot x^2$

$\frac {d}{dx} (\cos ^2(ln x)) = 2 \cos (ln x) (- \sin (ln x)) \frac{1}{x}$
$= - \frac{2 \sin (ln x) \cos (ln x)}{x} = \frac{ - \sin (2 ln x)}{x}$

$\frac{d}{dx} (\sin \sqrt{x}) = \cos \sqrt{x} \frac{d}{dx} (\sqrt{x})$
$= \cos \sqrt{x} \frac{1}{2 \sqrt{x}}$

$y = \sqrt{e ^{\sqrt{x}}}$
$\Rightarrow \frac{dy}{dx} = \frac{1}{2 \sqrt{e ^{\sqrt{x}}}} \times e ^{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}$
$\Rightarrow \frac{dy}{dx} = \frac{e ^{\sqrt{x}}}{4 \sqrt{x e ^{\sqrt{x}}}}$

$y = \frac{logx}{x}$
$\Rightarrow \frac{dy}{dx} = \frac{\frac{1}{x} . x - \log x}{x^2} = \frac{ 1 - \log x}{x^2}$