Question

Let $f\left(x\right)=\underset{n\to \infty }{lim}{\tan }^{-1}\left(4n^2(1-\cos \frac{x}{n})\right)$ and $g\left(x\right)=\underset{n\to \infty }{lim}\frac{n^2}{2}\ln \cos \left(\frac{2x}{n}\right)$, then $\underset{x\to 0}{lim}\frac{e^{-2g\left(x\right)}-e^{f\left(x\right)}}{x^6}$

• A. $\frac{8}{3}$
• B. $\frac{7}{3}$
• C. $\frac{5}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{8}{3}$

$f\left(x\right)=\underset{n\to \infty }{lim}ta{n}^{-1}\left(4n^2(1-cos\frac{x}{n})\right)$
$\Rightarrow f\left(x\right)=ta{n}^{-1}\left({lim}_{n\to \infty }\left(4n^2(1-cos\frac{x}{n})\right)\right)$
Replace $n$ with $\frac{1}{h}$, where as $n\to \infty ,h\to 0$, we get
$\Rightarrow f\left(x\right)=ta{n}^{-1}\left(4{lim}_{h\to 0}\left(\frac{1-cos\left(xh\right)}{h^2}\right)\right)$
$\because {lim}_{x\to 0}\frac{1-cosx}{x^2}=\frac{1}{2}$
$\Rightarrow f\left(x\right)=ta{n}^{-1}\left(4{lim}_{h\to 0}\left(x^2\frac{1-cos\left(xh\right)}{(xh{)}^2}\right)\right)$
$\Rightarrow f\left(x\right)=ta{n}^{-1}\left(2x^2\right)$
$g\left(x\right)=\underset{n\to \infty }{lim}\frac{n^2}{2}\ln \cos \left(\frac{2x}{n}\right)$
Replace $n$ with $\frac{1}{h}$, where as $n\to \infty ,h\to 0$, we get
$\Rightarrow g\left(x\right)={lim}_{h\to 0}\frac{ln\cos \left(2xh\right)}{2h^2}$
$\because {lim}_{x\to 0}\frac{log(1+h)}{h}=1\&{lim}_{x\to 0}\frac{1-cosx}{x^2}=\frac{1}{2}$
$\Rightarrow g\left(x\right)={lim}_{h\to 0}-\frac{\left(\frac{ln(1-(1-\cos \left(2xh\right)\left)\right)}{-(1-\cos \left(2xh\right))}\right)\cdot (1-\cos \left(2xh\right))}{2h^2}$
$\Rightarrow g\left(x\right)={lim}_{h\to 0}-\frac{2x^2(1-\cos \left(2xh\right))}{4(xh{)}^2}=-x^2$
$\Rightarrow g\left(x\right)=-x^2$
Now,
$L:\underset{x\to 0}{lim}\frac{e^{-2g\left(x\right)}-e^{f\left(x\right)}}{x^6}=\underset{x\to 0}{lim}\frac{e^{2x^2}-e^{ta{n}^{-1}2x^2}}{x^6}$
Assume $ta{n}^{-1}2x^2=\theta$, then $x^2=\frac{tan\theta }{2}$, where as $x\to 0,\theta \to 0$
$\therefore L={lim}_{x\to 0}8\frac{e^{tan\theta }-e^{\theta }}{ta{n}^3\theta }=8{lim}_{x\to 0}{e}^{\theta }\frac{e^{tan\theta -\theta }-1}{ta{n}^3\theta }$
$\Rightarrow L=8{lim}_{x\to 0}\frac{e^{tan\theta -\theta }-1}{{\theta }^3}$
Using L'Hospital's rule, we get
$L=\frac{8}{3}$