Why isn't this limit equal to $0$? $f(2)=4$, $g(2)=9$, $f'(2)=g'(2)$. 
$ \displaystyle \lim_{x \to 2} \frac{ \sqrt{f(x)}-2} { \sqrt{g(x)}-2} $.
Why isn't this limit equal to $0$? Since $f$ and $g$ are differentiable at $x=2$, that means they are continuous, so you should be able to evaluate the limit by direct substitution. Thanks!
 A: Substitution is perfectly appropriate, since the denominator does not evaluate to $0$. But I think you substituted $2 = f(x)$ instead of using $f(2) = 4$, and likewise, you substituted $g(x) =2$ instead of $g(2) = 9$.
$$\lim_{x\to 2} \frac{\sqrt{f(x) - 2}}{\sqrt{g(x) - 2}} = \frac {\sqrt{4 - 2}}{\sqrt{9-2}} = \sqrt{\frac{2}7}$$
ADDED: Even with the correction in formatting, substitution is perfectly appropriate to use, again, since the denominator does not evaluate to /approach $0$:
$$\lim_{x\to 2} \frac{\sqrt{f(x)} - 2}{\sqrt{g(x)} - 2} = \frac {\sqrt{4} - 2}{\sqrt{9} -2} = \frac{0}{1} = 0$$
A: You can evaluate by direct substitution provided that the function in the denominator does not approach 0 or infinity.
I'm kind of suspecting there is a typo in your OP concerning the square roots, but even if a minor modification is necessary to what is actually there, this should address your question.
A: The conditions given don't make much sense unless the limit you have to compute is
$$
\lim_{x\to 2}\frac{\sqrt{f(x)}-2}{\sqrt{g(x)}-3}
$$
where direct substitution is impossible. In this case the data about the derivatives allow you to use L'Hôpital's theorem and do
$$
\lim_{x\to 2}\frac{\sqrt{f(x)}-2}{\sqrt{g(x)}-3}
=
\frac{\dfrac{f'(2)}{2\sqrt{f(2)}}}{\dfrac{g'(2)}{2\sqrt{g(2)}}}
=
\frac{f'(2)}{4}\frac{6}{g'(2)}
=\frac{3}{2}
$$
with the further assumption that $f'(2)=g'(2)\ne0$. If the common value of the derivatives at $2$ is zero you can't say anything about the limit with the available information.
If the limit is really
$$
\lim_{x\to 2}\frac{\sqrt{f(x)}-2}{\sqrt{g(x)}-2}
$$
then it is $0$ by substituting.
A: Im not really sure how you are finding 2 f(2)=4 and g(2)=9 so f(2)-2= 4-2 and g(2)=9 2-9=7 so you have $(\frac {2}{7})^{1/2}$ no?
A: As written, $\displaystyle\lim_{x \to 2}\dfrac{\sqrt{f(x)-2}}{\sqrt{g(x)}-2}=\frac {\sqrt 2}1$ by direct substitution.  It would be zero, also by direct substitution, if the numerator were $\sqrt {f(x)}-2$
