Here is the limit I am trying to do
$$ \lim\limits_{x \to \infty} \frac{x^2 + \mathrm{e}^{4x}}{2x- \mathrm{e}^x} $$
Now, here first, I am trying to identify the indeterminate form so that I can use L'Hospital's rule. Numerator tends to $\infty$ as $x \to \infty $. But the denominator tends to $ \infty - \infty$ as $ x \to \infty$. So, indeterminate form would be
$$ \frac{\infty}{\infty - \infty} $$
So, how to approach this problem here ?
The denominator is indeed of the indeterminate form $\infty-\infty$, but simplifying it we get $$2x-e^x = e^x\left(\frac{2x}{e^x}-1\right)$$ and since $$\lim_{x\to\infty}\frac{2x}{e^x}=0$$ (e.g. by L'Hopital's), we see that the denominator tends to $-\infty$, so you can proceed with L'Hopital's for the original fraction as usual, as it is of the indeterminate form $\frac{\infty}{-\infty}$.
For the denominator, you have to find the limit $\lim\limits_{x \to \infty} (2x-\mathrm{e}^x)$. You have noticed that both terms tend to $\infty$, so we need a more nuanced comparison. (The limit of "$\infty-\infty$", so to speak, could be any real number, or $\pm \infty$ e.g. take $(x+k)-x$ to get limit $k$, and $2x-x$ for $\infty$, and $x-2x$ for $-\infty$.)
Recall that
$$ e^x=1+x+\frac{x^2}{2}+\cdots+\frac{x^n}{n!}+\cdots$$
So $$ 2x-e^x=-1+x-\frac{x^2}{2}-\cdots-\frac{x^n}{n!}-\cdots$$
What is the limit of this as $x\to \infty$?
