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