# Evaluating limits...

• Jan 31st 2010, 10:02 PM
Neversh
Evaluating limits...
$\lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}$

I can't seem to evaluate it properly...

I don't know how to use the math tag tool so I had to copy and tinker with it until I had what I needed... The edited equation is what I'm having difficulty with.
• Jan 31st 2010, 10:09 PM
Drexel28
Quote:

Originally Posted by Neversh
Lim ( 1/(x([1 + x]^1/2) -1/x)
x->0

I can't seem to evaluate it properly...

Let's assume the limit exists and equals $L\ne 0$. Then, $\frac{1}{L}=\lim_{x\to0}\frac{1}{\frac{1}{x\sqrt{1 +x}}-\frac{1}{x}}=\lim_{x\to0}\frac{x\sqrt{1+x}\cdot x}{x-x\sqrt{1+x}}=\lim_{x\to0}\frac{x\sqrt{1+x}}{1-\sqrt{1+x}}$ $=\lim_{x\to0}\frac{x}{\frac{1}{\sqrt{1+x}}-1}=\frac{1}{f'(0)}$ where $f(x)=\frac{1}{\sqrt{1+x}}$
• Jan 31st 2010, 11:50 PM
General
Quote:

Originally Posted by Neversh
$lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}$

The answer is - 1/2. I can't seem to get this answer :(

Start by making a common denominator.
• Feb 1st 2010, 12:34 AM
lovemath
Quote:

Originally Posted by Neversh
$\lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}$

The answer is - 1/2. I can't seem to get this answer :(

$\lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}=\lim_{x\to0}{\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}}$
$=\lim_{x\to0}{\frac{(1-\sqrt{1+x})(1+\sqrt{1+x})}{(x\sqrt{1+x})(1+\sqrt{1 +x})}}$
$=\lim_{x\to0}{\frac{(-x)}{(x\sqrt{1+x})(1+\sqrt{1+x})}}$
$=\lim_{x\to0}{\frac{(-1)}{(\sqrt{1+x})(1+\sqrt{1+x})}}$
$=\frac{-1}{2}$