1. ## Limits

I'm having trouble with these limits. Any help would be appreciated.

$\displaystyle \lim_{x\to +\infty}\frac{\sqrt[3]{{x}^{5}-3x}}{{x}^{2}+4}$

$\displaystyle \lim_{x\to 0^{+}} \sqrt{\frac{x+1}{x}}$

2. Originally Posted by Alvy
I'm having trouble with these limits. Any help would be appreciated.

$\displaystyle \lim_{x\to +\infty}\frac{\sqrt[3]{{x}^{5}-3x}}{{x}^{2}+4}$

$\displaystyle \lim_{x\to 0^{+}} \sqrt{\frac{x+1}{x}}$
Observe:

$\displaystyle \frac{\sqrt[3]{{x}^{5}-3x}}{{x}^{2}+4}=\frac{x^{5/3}(1-3/x^4)}{x^2(1+4/x^2)}=\frac{(1-3/x^4)}{x^{1/3}(1+4/x^2)}$

Now you should be able to determine the limit.

CB

3. Originally Posted by Alvy
I'm having trouble with these limits. Any help would be appreciated.

$\displaystyle \lim_{x\to +\infty}\frac{\sqrt[3]{{x}^{5}-3x}}{{x}^{2}+4}$

$\displaystyle \lim_{x\to 0^{+}} \sqrt{\frac{x+1}{x}}$
For the second one:

$\displaystyle \sqrt{\frac{x+1}{x}} = \sqrt{\frac{\frac{x+1}{x}}{\frac{x}{x}}} = \sqrt{\frac{1+{\frac{1}{x}}}{1}}= \sqrt {1+{\frac{1}{x}}}$

4. Thank you very much.