L'Hospital's Rule to find limit I am asked to find
$\lim\limits_{x \rightarrow \infty} {\left(\frac{8x}{8x+4}\right)}^{5x}$.
Could anyone help me with figuring out how to start this problem? Thanks!
 A: \begin{align}
\lim_{x \to \infty} \bigl(\frac{8x}{8x+4}\bigr)^{5x} = \lim_{x \to \infty} \mathrm{e}^{\log{\bigl(\bigl(\frac{8x}{8x+4}\bigr)^{5x}\bigr)}} = \lim_{x \to \infty} \mathrm{e}^{5x \log{\bigl(\frac{8x}{8x+4}\bigr)}} = \mathrm{e}^{\lim_{x \to \infty} 5x \log{\bigl(\frac{8x}{8x+4}\bigr)}}
\end{align}
See if you can solve $\lim_{x \to \infty} 5x \log \bigl( \frac{8x}{8x+4} \bigr)$ using l'Hôpital's rule.
A: Any function of the form $f(x)^{g(x)}$ should first be transformed into
$$
\exp(g(x)\log f(x))
$$
when computing a limit. Then, since $\exp$ is continuous, we can just compute the limit of $g(x)\log f(x)$ and exponentiate the result.
So you want to compute
$$
\lim_{x\to\infty}5x\log\frac{8x}{8x+4}
$$
that's an indeterminate form $\infty\cdot 0$, because
$$
\lim_{x\to\infty}\frac{8x}{8x+4}=1
$$
Such a form should be brought to either $0/0$ or $\infty/\infty$; the rule of thumb is “leave the logarithm at the numerator”, so we first try
$$
\lim_{x\to\infty}\frac{\log\frac{8x}{8x+4}}{1/x}
$$
(we can reinsert the factor $5$ at the end. We can apply l'Hôpital's theorem, but it's better to rewrite the numerator as $\log(8x)-\log(8x+4)$:
$$
\lim_{x\to\infty}\frac{\log\frac{8x}{8x+4}}{1/x}=
\lim_{x\to\infty}\frac{\log(8x)-\log(8x+4)}{1/x}\overset{\text{(H)}}{=}
\lim_{x\to\infty}\frac{\dfrac{8}{8x}-\dfrac{8}{8x+4}}{-1/x^2}
$$
and the final expression becomes
$$
-x^2\left(\frac{1}{x}-\frac{2}{2x+1}\right)=-x^2\frac{2x+1-2x}{x(2x+1)}=
-\frac{x}{2x+1}
$$
and you should be able to carry on from here.
A: rewrite the given term in the form $e^\frac{\ln\left(\frac{8x}{8x+4}\right)}{\frac{x}{5}}$ and use the rulse of L'Hopital
