Question

Let $f\left(x\right)=\underset{n\to \infty }{lim}{\left(\cos \sqrt{\frac{x}{n}}\right)}^n$, $g\left(x\right)=\underset{n\to \infty }{lim}{(1-x+x.\sqrt[n]{ne})}^n$
$\underset{x\to 0}{lim}\frac{\ell \left(f\left(x\right)\right)}{\ell \left(g\left(x\right)\right)}$ is equal to

Answer 1

$f\left(x\right)=\underset{n\to \infty }{lim}{\left(\cos \sqrt{\frac{x}{n}}\right)}^n$

$g\left(x\right)=\underset{n\to \infty }{lim}{(1-x+x.\sqrt[n]{ne})}^n$

Taking $\log$ on both sides

$\ln f\left(x\right)=\underset{n\to \infty }{lim}\frac{\frac{-\sin \sqrt{\frac{x}{n}}}{\cos \sqrt{\frac{x}{n}}}.\frac{-1}{2\sqrt{\frac{x}{n}}}.\frac{x}{n^2}}{\frac{-1}{n^2}}$

$\ln f\left(x\right)=\underset{n\to \infty }{lim}\frac{-\sin \sqrt{\frac{x}{n}}.x}{\cos \sqrt{\frac{x}{n}}.2\sqrt{\frac{x}{n}}}$

$\ln f\left(x\right)=-\frac{1}{2}.\underset{n\to \infty }{lim}\left(\frac{1}{1}\right)x.\frac{\sin \sqrt{\frac{x}{n}}}{\sqrt{\frac{x}{n}}}$

$\ln f\left(x\right)=-\frac{1}{2}\underset{n\to \infty }{lim}\frac{\sin \sqrt{\frac{x}{n}}}{\sqrt{\frac{x}{n}}}.x$

$\ln f\left(x\right)=\frac{-x}{2}......(\because n⟶\infty \sqrt{\frac{x}{n}}⟶0\Rightarrow {lim}_{\sqrt{\frac{x}{n}}\to 0}\frac{\sin \sqrt{\frac{x}{n}}}{\sqrt{\frac{x}{n}}}=1)$

$\Rightarrow f\left(x\right)=e^{-x/2}\to 1$

Now $g\left(x\right)=\underset{n\to \infty }{lim}{(1-x+x\sqrt[n]{ne})}^n$

Taking $\ln$ on both sides,

$\Rightarrow \ln g\left(x\right)=\underset{n\to \infty }{lim}n\log (1-x+x.e^{1/n})$

$\Rightarrow \ln g\left(x\right)=\underset{n\to \infty }{lim}\frac{\frac{1}{(1-x+x.e^{1/n})}.(x.e^{1/n}).\frac{-1}{n^2}}{\frac{-1}{n^2}}$ (Applying L-Hospital rule)

$\Rightarrow \ln g\left(x\right)=\frac{\left(\frac{1}{1}\right).x.\left(1\right)}{1}$

$\Rightarrow \ln g\left(x\right)=x$

$\Rightarrow g\left(x\right)=e^x$

${lim}_{x\to 0}\frac{\ell \left(f\left(x\right)\right)}{\ell \left(g\left(x\right)\right)}={lim}_{x\to 0}\frac{\ell \left(e^{-x/2}\right)}{\ell \left(e^x\right)}=\frac{\frac{-x}{2}}{x}=\frac{-1}{2}$