Question

$\underset{x\to \infty }{lim}{(1-\frac{1}{3x})}^{5x}$

Answer 1

We have,

${lim}_{x\to \infty }{(1-\frac{1}{3x})}^{5x}$

Now,

Let, $L={lim}_{x\to \infty }{(1-\frac{1}{3x})}^{5x}$

Taking log both side and we get,

$\Rightarrow \log L=\log \left({lim}_{x\to \infty }{(1-\frac{1}{3x})}^{5x}\right)$

$={lim}_{x\to \infty }\left(\log {(1-\frac{1}{3x})}^{5x}\right)$

$={lim}_{x\to \infty }\left(5x\log (1-\frac{1}{3x})\right)$

$={lim}_{x\to 0}\left(\frac{\log (3-x)}{15x}\right)\left(\text{Note}\text{the}\text{change}\text{limit}\right)$

It is $\frac{0}{0}form$

Then, using L’ Hospital rule and we get,

$={lim}_{x\to 0}\frac{1}{15(3-x)}$

Taking limit and we get,

$=\frac{1}{15(3-0)}$

$=\frac{1}{45}$

Hence, this is the answer.