Limits of indeterminate Difference Quotients $$\lim \limits_{h \to 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}$$
I simplified, even made a table and came up with $-1/25$ but it appears to be incorrect, according to my submit module. 
If anyone has any tips or tricks on finding these limits, it would be greatly appreciated! I understand that the one with h and x will have to be in terms of x in some form or another. 
$$\lim \limits_{h \to 0} \frac{\frac{1}{x+h+3}-\frac{1}{x+3}}{h}$$
EDIT: What a typo, it is supposed to be subtraction on the top. 
I'm going to add the full original problem to make sure I didn't make any mistakes early on, so I know if the module has a wrong answer or not. 
The problem states:
$$f(x)=\frac{1}{x+3}$$
Compute the limit of the difference quotients $$\lim \limits_{h \to 0} \frac{f(2+h)-f(2)}{h}$$
 A: It looks to me as if you correctly calculated $$\begin{align*}
\lim\limits_{h\to 0}\frac{\frac{1}{5+h}-\frac{1}{5}}{h} &= \lim\limits_{h\to 0}\frac{5-(5+h)}{5h(5+h)}\\
&= \lim\limits_{h\to 0}\frac{-1}{5(5+h)}\\
&= \frac{-1}{25}.
\end{align*}$$ This is the problem that I’d have expected to see if you’re doing difference quotients, but if the problem was really to calculate $$\lim\limits_{h\to 0}\frac{\frac{1}{5+h}+\frac{1}{5}}{h},$$ as you’ve written it, then what you’ve done is indeed wrong: in that case the numerator approaches $\frac25$, so the limit is $\infty$ as $h \to 0^+$ and $-\infty$ as $h \to 0^-$.
Before I say anything about the second problem, then, let me ask: are you sure that you’ve copied it correctly? I’d expect a minus sign in the numerator.
A: From the way the problem is stated, I think you're supposed to do it by recognizing that the limit equals the derivative $f'(2)$.
(Or else, if this problem was given before talking about derivatives, it's a warm-up problem to get a feel for the kind of limits you will be doing when you get to derivatives...)
