Evaluating $\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}$ The question is to evaluate the given limit :
$$\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}$$
When I'm trying to evaluate this I'm getting $1$ as my answer but the answer given in the text is $0$.  I'm wondering how can this limit be equal to $0$.
I'm using $x = 3 + h$ (assuming $h$ is approaching zero) and now converting this limit in the form of $h$. Lim (h tends to zero) and substituting $x = 3 + h$. 
In this way numerator equals $h$ and now after rationalizing the denominator I will get $2h$ and the numerator will also become $2h$ both of them will cancel each other and the limit will be equal to $1$. 
I don't know where I'm going wrong. 
 Please edit if you can for proper reading because I don't know how to edit this in mathematical format. 
 A: We also have
\begin{align*}
\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}&=\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}\cdot\frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}}\\
&=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(x-2)-(4-x)}\\
&=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2x-6}\\
&=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)}\\
&=\lim_{x \to 3}\frac{\sqrt{x-2}+\sqrt{4-x}}{2}\\
&=\frac{\sqrt{3-2}+\sqrt{4-3}}{2}\\
&=\frac{\sqrt{1}+\sqrt{1}}{2}\\
&=\frac{2}{2}\\
&=1.
\end{align*}
A: Your working is correct. The answer is one. This is probably one of those cases where the textbook has a typo in it.
$$\lim_{x\to3}\frac{x+3}{\sqrt{x-2}-\sqrt{4-x}}=\lim_{h\to0}\frac{h}{\sqrt{1+h}-\sqrt{1-h}}$$
$$=\lim_{h\to0}\frac{h(\sqrt{1+h}+\sqrt{1-h})}{(\sqrt{1+h}-\sqrt{1-h})(\sqrt{1+h}+\sqrt{1-h})}$$
$$=\lim_{h\to0}\frac{h(\sqrt{1+h}+\sqrt{1-h})}{2h}$$
$$=\lim_{h\to0}\frac{(\sqrt{1+h}+\sqrt{1-h})}{2}$$
$$=\frac{1+1}{2}$$
$$=1$$
A: One thing that's wrong is the + sign in the numerator.  Most likely it should be a - sign as suggested by @Henry.
If I multiply the numerator and denominator by $\sqrt{x-2}+\sqrt{4-x}$ I get a difference of squares in the new denominator, whose value cancels the $x-3$ factor.  Then I do in fact get 1.  0 and 1 are next to each other on a QWERTY board and somebody's finger slipped.
