Evaluating the limit if it exists of: $\lim\limits_{h\to 0}\frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$ I'd like to evalute the limit of:
$$\lim\limits_{h\to 0}\frac{\quad\frac{1}{(x+h)^2} - \frac{1}{x^2}\quad}{h}.$$
if it exists. What throws me off is that there is another variable (x). How do I approach this using Limit Laws and rearranging the equation only?
What I know so far is that the numerator and denominator equal 0 if I take the limit now, so I need to somehow rearrange this.
Thanks.
 A: Because the limit is about what happens as $h$ approaches $0$, $x$ is playing the role of an "unknown constant". That is, even though we think of $x$ as a variable, for the purposes of this limit $x$ is actually constant, so you can treat it as a constant.
To compute the limit, try doing the algebra to turn this compound fraction into a simple fraction, and then see if there are things that cancel out to allow you to evaluate the limit.
(As you probably know, this limit is, by definition, "the derivative of $f(x) = \frac{1}{x^2}$").
A: Explicitly, for your case,
multiplying numerator and denominator by $x^2 (x+h)^2$,
$$\frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \frac{x^2-(x+h)^2}{h x^2 (x+h)^2}
= \frac{-2xh-h^2}{h x^2 (x+h)^2} = \frac{-2x-h}{x^2 (x+h)^2}$$
What happens as $h \to 0$?
A: Hint:  put the two fractions over a common denominator.  This will give an $h$ in the numerator that can cancel with the one in the denominator.
A: Consider the slightly more general problem where we replace $\rm\:x^2\:$ by a polynomial $\rm\:f(x)\:.\:$ Then  
$$\displaystyle\rm\ \frac{1/f(x+h) - 1/f(x)}{h}\ =\ \frac{f(x+h)-f(x)}{h}\ \frac{-1}{ f(x)\ f(x+h)}\ $$
By the Factor Theorem, $\rm\:h = (x+h)-x\:$ divides $\rm f(x+h)-f(x)\:,\:$ so this factor of $\rm\:h\:\:$ cancels with the $\rm\:h\:$ in the denominator, making the limit determinate. 
In fact, the limit as $\rm\:h\to 0\:$ of the LHS is by definition the first derivative of $\rm\:1/f(x)\:,\:$ and the limit of the first factor of the RHS is by definition the first derivative of $\rm\:f(x)\:.\:$ Thus we deduce the derivative rule for inverses = reciprocals: $\rm\:(1/f(x))'\: =\: -\: f{\:'}(x)/f(x)^2\:.$
