# Math Help - Some easy ->infinity limits I can't solve

1. ## Some easy ->infinity limits I can't solve

Hi,

I'm having some problems with calculating quite a few limits, so I guess there's a concept I'm not getting...

Fe:
$\lim_{x\to\infty}x^{1/x}$
Or:
$\lim_{x\to\infty}x^4e^{-x^2}$

$\lim_{x\to\infty}\frac{e^x}{\sqrt(x)}$

How should I solve these?

Thanks!

2. Originally Posted by superdl
Hi,

I'm having some problems with calculating quite a few limits, so I guess there's a concept I'm not getting...

Fe:
$\lim_{x\to\infty}x^{\frac {1}{x}}$
Or:
$\lim_{x\to\infty}x^4e^{-x^2}$

$\lim_{x\to\infty}\frac{e^x}{\sqrt{x}}$

How should I solve these?

Thanks!
first one is one == 1
(hint: it's $\infty ^0$ so u have to ... == $\displaystyle e ^{\lim_{x\to\infty}\frac {log(x)}{x}}$ and so on ... )

$\displaystyle \lim_{x\to\infty}x^{\frac {1}{2}} = 1$

second is zero (first use L'Hospitals rule .... )
(hint: after first use of rule you should get $\displaystyle \lim_{x\to\infty} \frac {2x^2}{e^{x^2}}$ and after second use you will get $\displaystyle \lim_{x\to\infty} \frac {2}{e^{x^2}}$ ..... )

$\displaystyle \lim_{x\to\infty}\frac{e^x}{\sqrt{x}}=0$

third one is infinity (first use L'Hospitals rule .... )
(hint: after use of rule you will get $\displaystyle \lim_{x\to\infty} 2 e^x \sqrt{x}$ .... )

$\displaystyle \lim_{x\to\infty}\frac{e^x}{\sqrt{x}} = \infty$

3. That's what I thought, but I also read that infinity^0 = undefined ?

4. I see you've edited your post.

The first one: could you elaborate a bit further please? :$I understand the second one. I was doing l'Hospitals, but I didn't notice I could rule out the 2x every time (stupid mistake) I also understand the third one. Thanks for your help so far! 5. Originally Posted by superdl I see you've edited your post. The first one: could you elaborate a bit further please? :$

I understand the second one. I was doing l'Hospitals, but I didn't notice I could rule out the 2x every time (stupid mistake)
I also understand the third one.

Thanks for your help so far!
when you get form like $\infty^0$ u can transform it like :

$\displaystyle \lim_{x\to \infty } x^{\frac{1}{x}} = e^{\displaystyle \lim_{x\to \infty }}\frac {log{(x)}}{x}}$

now u use L'Hospital's rule again and you'll get :

$\displaystyle e^{\displaystyle \lim_{x\to \infty } \frac {1}{x}} = 1$

6. I did some more researching and I get it know. Although, shouldn't it be ln instead of log ?

7. Originally Posted by superdl
Although, shouldn't it be ln instead of log ?
There seems to be a trend, particularly in US mathematics, to use $\log(x)$ in place of $\ln(x)$.
The late Leonard Gillman (Rings of Continuous Functions) and past president of the MAA was most vocal about this.

8. Do you mean that log should be used as the notation for the logarithm with base e then?

'cause $e^{ln a} = a$, and $10^{log a} = a$, but $e^{log a} \not= a$

(Ps: As you may have noticed, I'm not from the US; I live in Belgium)