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.

Note that $$2e^{2x}-e^{4x}=-e^{4x}(1-2e^{-2x})$$ – Mark Viola Feb 08 '18 at 20:09 — Mark Viola, Feb 08 '18 at 20:09

score 3 · Accepted Answer · answered Feb 08 '18 at 18:59

3

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 $-\infty$.

answered Feb 08 '18 at 18:59

vadim123

It might be better to write $2e^{2x}-e^{4x}=-e^{4x}(1-2e^{-2x})$ – Mark Viola Feb 08 '18 at 20:09

score 0 · Answer 2 · answered Feb 08 '18 at 19:01

0

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.

answered Feb 08 '18 at 19:01

user

2 Answers2