Simple limit, wolframalpha doesn't agree, what's wrong? (Just the sign of the answer that's off) $\begin{align*}
 \lim_{x\to 0}\frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}
 &=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{4+x}}-\frac{\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
 &=\lim_{x\to 0}\frac{\frac{2-\sqrt{4+x}}{2\sqrt{4+x}}}{x}\\
 &=\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}}\\
 &=\lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}\\
 &=\lim_{x\to 0}\frac{2 \times 2 + 2\sqrt{4-x}-2\sqrt{4-x}-((\sqrt{4-x})(\sqrt{4-x})) }{2 \times 2x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\
 &=\lim_{x\to 0}\frac{4-4+x}{4x\sqrt{4+x} + 2x\sqrt{4+x}\sqrt{4-x}}\\
 &=\lim_{x\to 0}\frac{x}{x(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\
 &=\lim_{x\to 0}\frac{1}{(4\sqrt{4+x} + 2\sqrt{4+x}\sqrt{4-x})}\\
&=\frac{1}{(4\sqrt{4+0} + 2\sqrt{4+0}\sqrt{4-0})}\\
&=\frac{1}{16} 
\end{align*}$
wolframalpha says it's negative. What am I doing wrong? 
 A: Others have already pointed out a sign error. One way to avoid such is to first simplify the problem by changing variables. Let $\rm\ z = \sqrt{4+x}\ $ so $\rm\ x = z^2 - 4\:.\:$ Then
$$\rm \frac{\frac{1}{\sqrt{4+x}}-\frac{1}{2}}{x}\ =\ \frac{\frac{1}z - \frac{1}2}{z^2-4}\ =\ \frac{-(z-2)}{2\:z\:(z^2-4)}\ =\ \frac{-1}{2\:z\:(z+2)}$$
In this form it is very easy to compute the limit as $\rm\ z\to 2\:$.
A: Certainly for $x \gt 0,\frac{1}{\sqrt{4+x}}-\frac{1}{2} \lt 0$ so the limit should be negative.  Between the  fifth and sixth limit you flipped a sign under the sqrt in the numerator and that changes the sign of the total thing
A: In between the fourth and fifth steps, you go from 
$$\lim_{x\to 0}\frac{2-\sqrt{4+x}}{2x\sqrt{4+x}} \text{ to } \lim_{x\to 0}\frac{(2-\sqrt{4-x})(2+\sqrt{4-x})}{(2x\sqrt{4+x})(2+\sqrt{4-x})}$$
which is not correct. It should be 
$$\lim_{x\to 0}\frac{(2-\sqrt{4+x})(2+\sqrt{4+x})}{(2x\sqrt{4+x})(2+\sqrt{4+x})}$$
A: An argument why it should be negative.
When $x>0$, we have $\sqrt{4+x} > 2$ and hence $\frac{1}{\sqrt{4+x}} < \frac{1}{2}$ and hence
$$\frac{\frac{1}{\sqrt{4+x}} - \frac{1}{2}}{x} < 0$$
Similarly, When $x<0$, we have $\sqrt{4+x} < 2$ and hence $\frac{1}{\sqrt{4+x}} > \frac{1}{2}$ and hence
$$\frac{\frac{1}{\sqrt{4+x}} - \frac{1}{2}}{x} < 0$$
So either way, as $x \rightarrow 0$, the value is negative.
