Limit as x goes to infinity with logarithm in the exponent I try to solve this limit without results, could someone help me?
$$ \lim_{x\to\infty} \biggl(\frac {e^x-x^2 }{e^x-5x}\biggr)^{\log x} $$
 A: It is enough to find the limit of 
$$\log  x\,\bigl(\log(1-x^2\mathrm e^{-x})-\log(1-5x\mathrm e^{-x})\bigr)$$
We'll use an asymptotic expansion. As $x^a\mathrm e^{-x}$ tends to $0$ as $x$ tends to $+\infty$, we have


*

*$\log(1-x^2\mathrm e^{-x})=-x^2\mathrm e^{-x}+o(x^2\mathrm e^{-x})$

*$\log(1-5x\mathrm e^{-x})=-5x\mathrm e^{-x}+o(x\mathrm e^{-x})$,


so that
\begin{align*}
\log(1-x^2\mathrm e^{-x})-\log(1-5x\mathrm e^{-x})&=(5x-x^2)\mathrm e^{-x}+o(x^2\mathrm e^{-x})+o(x\mathrm e^{-x})\\
&=-x^2\mathrm e^{-x}+o(x^2\mathrm e^{-x})\sim_\infty-x^2\mathrm e^{-x}
\end{align*}
and finally
$$\log x\,\bigl(\log(1-x^2\mathrm e^{-x})-\log(1-5x\mathrm e^{-x})\bigr)\sim_\infty -x^2\mathrm e^{-x}\log x=o(x^3\mathrm e^{-x})\xrightarrow[x\to+\infty]{} 0.$$
A: Whenever we need to evaluate the limit of an expression of type $\{f(x)\}^{g(x)}$ then the best approach is to take logarithms.
Thus if $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{x \to \infty}\left(\frac{e^{x} - x^{2}}{e^{x} - 5x}\right)^{\log x}\right\}\notag\\
&= \lim_{x \to \infty}\log\left(\frac{e^{x} - x^{2}}{e^{x} - 5x}\right)^{\log x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to \infty}\log x\log\left(1 + \frac{e^{x} - x^{2}}{e^{x} - 5x} - 1\right)\notag\\
&= \lim_{x \to \infty}\log x\cdot\dfrac{5x - x^{2}}{e^{x} - 5x}\cdot\dfrac{\log\left(1 + \dfrac{5x - x^{2}}{e^{x} - 5x}\right)}{\dfrac{5x - x^{2}}{e^{x} - 5x}}\notag\\
&= \lim_{x \to \infty}\frac{\log x}{x}\cdot\dfrac{5x^{2} - x^{3}}{e^{x} - 5x}\notag\\
&= 0\cdot 0 = 0\notag
\end{align}
Hence $L = 1$. We have used the standard limits $$\lim_{x \to \infty}\frac{\log x}{x} = \lim_{x \to \infty}\frac{x}{e^{x}} = 0\tag{1}$$ which easily leads to $$\lim_{x \to \infty}\frac{5x - x^{2}}{e^{x} - 5x} = \lim_{x \to \infty}\frac{5x^{2} - x^{3}}{e^{x} - 5x} = 0\tag{2}$$
A: Here is a hand-wavey reasoning that tells you that if there is any justice in this world, then the limit should be $1$. This is not a proof, just strong evidence (it could probably be turned into a proof with the right tools).
First off, it is known that $\lim_{x\to\infty}(1-\frac{1}x)^x=\frac 1e$. This demonstrates that if the limit of an expression like $(1+\frac{1}{f(x)})^{g(x)}$ is to have a limit other than $0,1$ or $\infty$, then $f$ and $g$ should have the same order.
Your expression can be rewritten as
$$
\lim_{x\to\infty}\left(1-\frac{1}{(e^x-5x)/(5x-x^2)}\right)^{\log x}
$$and we see that $f$ is (approximately) exponential while $g$ is logarithmic. That's quite a big difference in order, which strongly suggests that the limit is indeed $1$.
