# Math Help - Proving Limits

1. ## Proving Limits

I need to prove the following statement

lim
x---> Infinity

3x^5-5logex/5e^x-2x^5

I have

logex/x^5 --> 0 as x---> Infinity

not sure about the other bit.

Also need to prove the following

lim
x--->o

x^2/1-e^x2 = -1

Can anyone help?

2. ## Re: Proving Limits

Almost impossible to make any sense of that. Please remember, if nothing else, that the Order of Operations still works.

These are very different animals:

x^2 / 1 - e^x2

and

x^2 / (1 - e^x2)

Also, what does "e^x2" mean? Maybe $e^{x^{2}}$? Or something else?

3. ## Re: Proving Limits

Try to multiply top and bottom by $e^{-x}$ and observe what it happens... and also try to learn latex and observe what it happens ...

Kind regards

$\chi$ $\sigma$

4. ## Re: Proving Limits

Thanks. I have a problem with advance settings, I can get the characters up to place them in the tex. Can anyone help?

5. ## Re: Proving Limits

Sorry I do mean e^x^2 and without the brackets.

6. ## Re: Proving Limits

You have to calculate:
$\lim_{x \to 0} \frac{x^2}{1-e^{x^2}}=-1$
By using l'Hopital's rule we get:
$\lim_{x \to 0} \frac{2x}{-2x\cdot e^{x^2}}=\lim_{x \to 0} \frac{1}{-e^{x^2}}=\frac{1}{-1}=-1$

7. ## Re: Proving Limits

The first limit:
$\lim_{x\to \infty}\frac{3x^5-\log(e^x)}{5e^x-2x^5}=...?$

Can we assume $\log(e^x)$ is the natural logarithm?
(I'm not from the USA, so I don't know if they mean the natural logaritm or the logarithm with base 10, because I always use $\ln$ for the natural logarithm)

yes.

9. ## Re: Proving Limits

Ok, maybe it can be useful to wright:
$5\log(e\cdot x)=5\log(e)+5\log(x)=5+5\log(x)$

10. ## Re: Proving Limits

ok, but how do we get to the limit = 0

11. ## Re: Proving Limits

Originally Posted by Arron
ok, but how do we get to the limit = 0
Read the replies given to you! You were told how to do this limit in post #6. If you need more help with it, quote the post and say what you still don't understand.

12. ## Re: Proving Limits

To prove that:
$\lim_{x \to \infty} \frac{3x^5-5\log(e\cdot x)}{5\cdot e^x-2x^5}=0$
My solution would be:
$\lim_{x\to \infty}\frac{3x^5-5\log(e\cdot x)}{5\cdot e^x-2x^5}=\lim_{x\to \infty} \frac{3-\frac{5\log(e\cdot x)}{x^5}}{\frac{5\cdot e^x}{x^5}-2}$

In general:
$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x \to a} f(x)}{\lim_{x\to a}g(x)}$

And so:
$\frac{3-\lim_{x\to \infty }\frac{5\log(e\cdot x)}{x^5}}}{\lim_{x\to \infty}\frac{5\cdot e^x}{x^5}-2}}$

We can calculate now the two limits by using l'Hopitals rule:
$\lim_{x\to \infty}\frac{5\log(e\cdot x)}{x^5}=0$
$\lim_{x\to \infty}\frac{5\cdot e^x}{x^5}=\infty$

So that means we get:
$\lim_{x \to \infty} \frac{3}{\infty}=0$