Question

Evaluate the limit as x goes to infinity using L'Hospital's rule:

8xe^(1/x)-8x

Answer 1

First note that \(\large \displaystyle \lim_{x\to\infty} 8xe^{1/x} - 8x = \lim_{x\to\infty} 8x(e^{1/x}-1)\rightarrow \infty\cdot 0\).

In order to apply L'Hopital's rule, you need to rewrite the limit so that it approaches either \(\large \dfrac{\infty}{\infty}\) or \(\large \dfrac{0}{0}\). In this case, it's easier to get it into the form \(\large \dfrac{0}{0}\) by seeing that
\[\large \lim_{x\to\infty} 8x(e^{1/x}-1) = \lim_{x\to\infty} \frac{8(e^{1/x}-1)}{\dfrac{1}{x}}\rightarrow \frac{0}{0}.\]
Now apply L'Hopital's rule to get \[\large \lim_{x\to\infty} \frac{8(e^{1/x}-1)}{\dfrac{1}{x}} = \lim_{x\to\infty} \frac{\dfrac{d}{dx}\left(8(e^{1/x}-1)\right)}{\dfrac{d}{dx}\left(\dfrac{1}{x}\right)}=\ldots\] Can you finish this off? Does this make sense? :-)

Answer 2

Yep! I found the right answer but does every problem eventually give a numerical answer if you use L'Hopital's Rule?

Answer 3

I'm pretty sure it should, given that the limit you apply L'Hopital's rule to is of the form \(\large \dfrac{\infty}{\infty}\) or \(\large \dfrac{0}{0}\).

Each limit that approaches one of the indeterminate values \(\large 1^{\infty},\,\infty^0,\, 0^0, \infty\cdot 0, \infty-\infty\) can be transformed into a limit of the form \(\large \dfrac{\infty}{\infty}\) or \(\large \dfrac{0}{0}\). Out of these, the cases you should be most careful with are \(\large \infty^0\) and \(\large \infty\cdot 0\); sometimes, if you express them in the form \(\large \dfrac{\infty}{\infty}\), you might see that you're not getting anywhere with it (no matter how many times you apply L'Hopital's rule), but if you wrote it in the form \(\large \dfrac{0}{0}\), then you get somewhere (and vice versa).

Other than that, I'm pretty sure you'll get a numerical limit whenever you apply L'Hopital's rule.

I hope this clarifies things! :-)