<s>[INST] How to evaluate $\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$
 [/INST] With limits going to infinity, usually factoring out the greatest term is the way to go, since then you'll have a lot of terms going to $0$. Here the term that grows faster is $x!$. So:
\begin{align}
\require{cancel}
\lim_{x \to +\infty} \frac{(x+2)(x+1)x!\left(1+\frac{4^x}{(x + 2)!}\right)}{((2x+1)^2+\ln x)x!} &= \lim_{x \to +\infty} \frac{(x^2+3x+2)\left(1+\frac{4^x}{(x + 2)!}\right)}{4x^2\left(1 + \frac1x + \frac1{4x^2}+\frac{\ln x}{4x^2}\right)} =\\
&= \lim_{x \to +\infty} \frac{\cancel{x^2}\left(1+\frac3x+\frac2{x^2}\right)\left(1+\frac{4^x}{(x + 2)!}\right)}{4\cancel{x^2}\left(1 + \frac1x + \frac1{4x^2}+\frac{\ln x}{4x^2}\right)} = \frac14.
\end{align}
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