1. ## Evaluating limits...

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

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.

2. 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 $\displaystyle L\ne 0$. Then, $\displaystyle \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}}$$\displaystyle =\lim_{x\to0}\frac{x}{\frac{1}{\sqrt{1+x}}-1}=\frac{1}{f'(0)}$ where $\displaystyle f(x)=\frac{1}{\sqrt{1+x}}$

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

The answer is - 1/2. I can't seem to get this answer
Start by making a common denominator.

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

The answer is - 1/2. I can't seem to get this answer
$\displaystyle \lim_{x\to0}{\frac{1}{x\sqrt{1+x}}-\frac{1}{x}}=\lim_{x\to0}{\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}}$
$\displaystyle =\lim_{x\to0}{\frac{(1-\sqrt{1+x})(1+\sqrt{1+x})}{(x\sqrt{1+x})(1+\sqrt{1 +x})}}$
$\displaystyle =\lim_{x\to0}{\frac{(-x)}{(x\sqrt{1+x})(1+\sqrt{1+x})}}$
$\displaystyle =\lim_{x\to0}{\frac{(-1)}{(\sqrt{1+x})(1+\sqrt{1+x})}}$
$\displaystyle =\frac{-1}{2}$