Evaluating Limits

**CrazyCat87** · February 10th 2010, 07:14 AM

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

**felper** · February 10th 2010, 07:21 AM

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.

**CrazyCat87** · February 10th 2010, 07:32 AM

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!

**HallsofIvy** · February 10th 2010, 07:49 AM

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

**CrazyCat87** · February 10th 2010, 07:50 AM

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

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