1. ## Find the Limits

I'm pretty sure i do something wrong here, i get both limits to 0. Any help would be greatly appreciated!
$\displaystyle \lim\frac{4n+1}{\sqrt{4n^2-3}}$ and $\displaystyle \lim\sqrt{n^2-3n}-n$

n->infinity (for both, couldnt figure out how to add it to the lim with the latex)

2. Originally Posted by takamura
I'm pretty sure i do something wrong here, i get both limits to 0. Any help would be greatly appreciated!
$\displaystyle \lim\frac{3n+2}{\sqrt{4n^2-1}}$ and $\displaystyle \sqrt{n^2-5n}-n$

n->infinity (for both, couldnt figure out how to add it to the lim with the latex)
What value of x are you making the function tend to?

3. Originally Posted by takamura
$\displaystyle \lim_{n\rightarrow\infty}\frac{4n+1}{\sqrt{4n^2-3}}$
Try dividing, top and bottom, by n. (Inside the square root, of course, this will turn into n^2.)

Originally Posted by takamura
$\displaystyle \lim_{n\rightarrow\infty}\sqrt{n^2-3n}-n$
Try putting this over "1", and multiplying, top and bottom, by the conjugate of the original expression. Then divide, top and bottom, by n.