L Hospital Rule

18.

(Infinity/Infinity Form)

A.L.H.R,
(sec²x/tanx)/(1/x)

xsec²x/tanx (0/0 Form)

A.L.H.R,

(x2sec²xtanx + sec²x)/sec²x

Substituting 0,

Answer:1

\hspace{-16}\bf{(19)\;}$ Given $\bf{\lim_{x\rightarrow 0}\frac{\ln(\tan 2x)}{\ln(\tan x)}}$\\\\\\ Now when $\bf{x\rightarrow 0\;,}$ Then $\bf{\tan (2x)\approx 2x}$\\\\\\ and $\bf{\tan (x)\approx x}$.\\\\\\ So $\bf{\lim_{x\rightarrow 0}\frac{\ln(\tan 2x)}{\ln(\tan x)}\approx \lim_{x\rightarrow 0}\frac{\ln(2x)}{\ln(x)}\Rightarrow _{Using \; L.Hopital \; Rule}\frac{2}{2x}\cdot \frac{x}{1}=1}$

\hspace{-16}\bf{(13)\;}$ Given $\bf{\lim_{x\rightarrow \frac{1}{2}}\frac{\cos^2(\pi x)}{e^{2x}-2e.x}=\frac{1}{2}\lim_{x\rightarrow \frac{1}{2}}\frac{2\cos^2 (\pi x)}{e^{2x}-2e.x}}$\\\\\\ $\bf{=\frac{1}{2}\lim_{x\rightarrow \frac{1}{2}}\frac{1+\cos (2\pi x)}{e^{2x}-2e.x}\Rightarrow_{Using \; L.Hopital\; Rule} \frac{1}{2}\lim_{x\rightarrow \frac{1}{2}}\frac{-2\pi.\sin (2\pi x)}{2e^{2x}-2e}}$\\\\\\ $\bf{=\Rightarrow_{Using \; L.Hopital\; Rule} \frac{1}{2}\lim_{x\rightarrow \frac{1}{2}}\frac{-4\pi^2.\cos (2\pi x)}{4e^{2x}}=\frac{1}{2}.\frac{4\pi^2}{4e}=\frac{\pi^2}{2e}}$