Evaluating limits...

**Neversh** · January 31st 2010, 09:02 PM

$\lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}$

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.

**Drexel28** · January 31st 2010, 09:09 PM

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 $f(x)=\frac{1}{\sqrt{1+x}}$

**General** · January 31st 2010, 10:50 PM

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.

**lovemath** · January 31st 2010, 11:34 PM

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

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

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