# Thread: Limits question

1. ## Limits question

Find the limit or show that no limit exists. I didn't really understand the lecture but I think the limit doesn't exist here. Can anyone help? Thanks.

2. Originally Posted by Thomas154321

Find the limit or show that no limit exists. I didn't really understand the lecture but I think the limit doesn't exist here. Can anyone help? Thanks.
$\lim_{x \to \infty} \frac{n^2 + log(n)}{\sqrt{2n^3 - 1}}$

First note that
$\lim_{n \to \infty} \frac{log(n)}{\sqrt{2n^3 - 1} }= 0$ <-- You show this.
so we don't have to worry about that term.

Thus
$\lim_{x \to \infty} \frac{n^2 + log(n)}{\sqrt{2n^3 - 1}} = \lim_{x \to \infty} \frac{n^2}{\sqrt{2n^3 - 1}}$

$= \lim_{x \to \infty} \frac{n^2}{\sqrt{2}n^{3/2}} = \frac{1}{\sqrt{2}} \lim_{x \to \infty} n^{1/2} \to \infty$

-Dan

3. Thanks for the fast reply!

Just wondering why you can just ignore the 1 in the denominator in the last line.

This was another one:

I think the answer is -2 but not sure, can you check please? Thanks so much!

4. Originally Posted by Thomas154321
Thanks for the fast reply!

Just wondering why you can just ignore the 1 in the denominator in the last line.
Think of it this (un-math-proof-like) way: As n gets very very large, how does $2n^2$ compare to that measly 1?

I'll leave it to you to formalize this statement.

-Dan

5. Originally Posted by Thomas154321

$\lim_{n \to \infty}n^2 \left ( \frac{1}{n + 1} - \frac{1}{n - 1} \right ) = \lim_{n \to \infty} n^2 \cdot \frac{-2}{(n + 1)(n - 1)}$

$= \lim_{n \to \infty} \frac{-2n^2}{n^2 - 1}$

$= \lim_{n \to \infty} \frac{-2n^2}{n^2 - 1} \cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}}$

$= \lim_{n \to \infty} \frac{-2}{1 - \frac{1}{n^2}} = -2$

-Dan