# Determine the asymptotes of the function $f(x)=x(2^{1/x}-2^{-1/x})$

Determine the asymptotes of the following function: $$f(x)=x(2^{1/x}-2^{-1/x})$$ So far, I have tried calculating the limits of the function at 0(the only point at which the function isn't defined) in order to check for the existence of a vertical asymptote at that point, and the limits at negative and positive infinity in order to determine the presence of horizontal asymptotes.

However, these limits ended up being of the forms zero times infinity and infinity times infinity, neither of which I know how to transform into a form which I would be able to evaluate in order to get the results, which are $x=0$ and $y=\ln{4}$ according to Wolfram Alpha. I also wasn't able to find any similar examples in the literature and lecture notes I'm using.

• I'm afraid you'll have to do more than that, if you're looking for any kind of help here. You should expand your post in terms of context and attempts at a solution. This site is not a homework service – Victor Oct 4 '15 at 19:27
• Please show your effort or what you have tried. :) – H. R. Oct 4 '15 at 19:34
• Have you tried L'Hopital's rule? – Paul Sinclair Oct 4 '15 at 20:32
• @PaulSinclair I've seen it mentioned in a couple of places, but I assumed it doesn't apply to this task as it won't be covered for two more weeks as part of the course I got this assignment from. I'll take a look at it. – akukas Oct 4 '15 at 20:47
• The problem is, we don't know what tools are currently available to you. The $x = 0$ asymptote is fairly easy to show by various approximations. But while $\ln 4$ drops out very easily by L'Hopital, demonstrating it by other means requires some finesse. – Paul Sinclair Oct 4 '15 at 20:53

$\lim_{x \to \infty} f = \frac{2^{1/x} - 2^{-1/x}}{1/x}$
Replace $x$ with $1/t$ and using L'Hopital's rule:
$\lim_{t \to 0} f = \frac{2^t - 2^{-t}}{t} = \log 2 \lim_{t \to 0} \frac{2^t + 2^t}{1} = 2 \log 2 = \log 4$