# Math Help - Limit x->0+ X^(2/x) HELP!

1. ## Limit x->0+ X^(2/x) HELP!

I need know how to show the work on this limit problem. I have been to several school tutors who can't seem to do it correctly. What is the limit as x approaches 0 from the right of x^(2/x)? The question as you would type it into an Computer Algebra System is: limit(x^(2/x),x,0,1)
This gives us 0 which is the correct answer but I have no idea how to show the work. I know that in the end it someone should be e^(-∞) which equals 0. Someone please help!

2. ## Re: Limit x->0+ X^(2/x) HELP!

Have you considered the logarithm? $log\left(x^{\frac{2}{x}}\right) = \frac{2}{x}\cdot log(x)$

3. ## Re: Limit x->0+ X^(2/x) HELP!

Yeah I got there. It makes sense now. I guess it ends up being -∞ because 2ln(.0000000000...1) is a large negative number over a very very small number. Thank you for the reminder. I made the mistake of trying to use L'Hospitals rule since that was the chapter this questions was in :/.
Thanks!

4. ## Re: Limit x->0+ X^(2/x) HELP!

Do you know L'Hôpital's rule?

Write $\displaystyle x^{2/x}\text{ as }\left(e^{\ln(x)\right) ^{2/x}=\left(e\right)^{\displaystyle\frac{2\ln(x)}{x}}$

The exponential function is continuous, so $\lim_{x\to a}(e^{f(x)})=\left(e\right)^{ \lim_{x\to a}(f(x))}$

Use L'Hôpital's rule to evaluate the limit, $\lim_{x\to0^+}\frac{2\ln(x)}{x}}$

5. ## Re: Limit x->0+ X^(2/x) HELP!

I didn't think I could use L'Hopital's rule because I can't seem to get the correct indeterminate form (∞/∞, -∞/-∞, ect..).

6. ## Re: Limit x->0+ X^(2/x) HELP!

Originally Posted by NeedDirection
I didn't think I could use L'Hopital's rule because I can't seem to get the correct indeterminate form (∞/∞, -∞/-∞, ect..).
Doh!

Yup, that's $\frac{=\infty}{0^+}$, which is -∞ , so that's good!