# Evaluating Limits

• Feb 10th 2010, 07:14 AM
CrazyCat87
Evaluating Limits
I need to evaluate the following limits:
1) $\lim_{x\to0} (\sqrt{x+1})/x$, where $(x>-1)$
2) $\lim_{x\to\infty} (\sqrt{x+1})/x$, where $(x>0)$

here is what I have so far
$(\sqrt{x+1})/x = \sqrt{\frac{x+1}{x^2}} = \sqrt{\frac{1}{x} + \frac{1}{x^2}} < \sqrt{\frac{2}{x}}$ . . . I don't know if I'm on the right track or not. Any help is greatly appreciated,thanks
• Feb 10th 2010, 07:21 AM
felper
Quote:

Originally Posted by CrazyCat87
I need to evaluate the following limits:
1) $\lim_{x\to0} (\sqrt{x+1})/x$, where $(x>-1)$
2) $\lim_{x\to\infty} (\sqrt{x+1})/x$, where $(x>0)$

here is what I have so far
$(\sqrt{x+1})/x = \sqrt{\frac{x+1}{x^2}} = \sqrt{\frac{1}{x} + \frac{1}{x^2}} < \sqrt{\frac{2}{x}}$ . . . I don't know if I'm on the right track or not. Any help is greatly appreciated,thanks

$= \sqrt{\frac{1}{x} + \frac{1}{x^2}}$

Using that you can conclude. If x is infinity, your fractions will be 0. If x is 0, your fractions will be infinitly big, infinity.
• Feb 10th 2010, 07:32 AM
CrazyCat87
Quote:

Originally Posted by felper
$= \sqrt{\frac{1}{x} + \frac{1}{x^2}}$

Using that you can conclude. If x is infinity, your fractions will be 0. If x is 0, your fractions will be infinitly big, infinity.

Yea I undertsand that but I am required to prove these statements strictly by definition. Otherwise it'd be easy!
• Feb 10th 2010, 07:49 AM
HallsofIvy
Then you should have said that! What you said was "I need to evaluate the following limits".
• Feb 10th 2010, 07:50 AM
CrazyCat87
Quote:

Originally Posted by HallsofIvy
Then you should have said that! What you said was "I need to evaluate the following limits".

Yea that's my fault, our ta says that but expects us to have a full proof lol