Find the value of $\lim _{x\to \infty }\left(1+\frac{4}{\ln x}\right)^{\ln\left(x^2+x\right)}$ So this is my attempt to the question:
Let,
$$y=\left(1+\frac{4}{\ln x}\right)^{\ln\left(x^2+x\right)}$$
$$\ln y=\ln(x^2+x)\ln\left(1+\frac{4}{\ln x}\right)$$
$$\lim _{x\to \infty }\ln y=\lim _{x\to \infty }\ln(x^2+x)\ln\left(1+\frac{4}{\ln x}\right)$$
Solving for the right side, since it is in the form of $\infty\cdot0$, we can change this into
$$\lim _{x\to \infty }\frac{\ln(x^2+x)}{\frac{1}{\ln\left(1+\frac{4}{\ln x}\right)}}$$
So that we can then use the L'Hopital rule.
However, after using the L'Hopital rule once, I found the form to be even more complicated and still becomes an indeterminate form.
Is there another way to solve this problem?
 A: You can observe that,
$$
\begin{align}\lim _{x\to \infty }\left(1+\frac{4}{\ln x}\right)^{\ln\left(x^2+x\right)}=\lim _{x\to \infty }\left(\left(1+\frac{4}{\ln x}\right)^{\frac {\ln x}{4}}\right)^{\frac {4\ln (x^2+x)}{\ln x}}\end{align}
$$
Since $\lim _{x\to \infty }\left(1+\frac{4}{\ln x}\right)^{\frac {\ln x}{4}}=e$, we need to evaluate the limit $$\lim_{x\to\infty }\frac {\ln (x^2+x)}{\ln x}.$$
Using the following inequality
$$
\begin{align}2=\frac {\ln (x^2)}{\ln x}≤\frac {\ln (x^2+x)}{\ln x}≤\frac {\ln (x^2+x^2)}{\ln x}=2+\frac {\ln 2}{\ln x}\end{align}
$$
and applying the Squeeze theorem we have:
$$\lim_{x\to\infty}\frac {\ln (x^2+x)}{\ln x}=2.$$
Thus, we conclude:
$$\begin{align}\lim _{x\to \infty }\left(1+\frac{4}{\ln x}\right)^{\ln\left(x^2+x\right)}&=e^{4\cdot 2}\\
&=e^8.\end{align}$$
A: $L=\lim\limits_{x\to \infty}\left(1+\frac{4}{\ln{x}}\right)^{\ln{(x^2+x)}}=\lim\limits_{x\to \infty}\left(1+\frac{4}{\ln{x}}\right)^{\ln{x}}\left(1+\frac{4}{\ln{x}}\right)^{\ln{(x+1)}}$
Let $y=\ln{x}$.
$L=\lim\limits_{y\to\infty}\left(1+\frac{4}{y}\right)^y \left(\left(1+\frac{4}{y}\right)^y\right)^{\ln{(x+1)}/\ln{x}}=e^4\left(e^4\right)^1=e^8$
