L'Hospital's rule with the indeterminate form of infinity minus infinity

Question

I am having trouble finding a way to take $\lim\limits_{x \to \infty} 2e^{2x} - e^{4x}$ with L'Hospital's rule.

Answer 1

There is no need for L'Hopital's rule:

$$\lim_{x\to\infty} e^{2x} (2-e^{2x})$$

This is of the form $(\infty)(-\infty)$, whose limit is $-\infty$.

Answer 2

You don’t need l’Hopital, simply observe that

$$2e^{2x} - e^{4x}=e^{2x}(2 - e^{2x})\to -\infty$$

which is obvious since $-e^{4x}$ dominates as $x$ increases.