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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 demoninator equal 0 if I take the limit now, so I need to somehow rearrange this.

Thanks.

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For the title, it's better to keep the equation "in-line" rather than displayed. The niceties (adding the extra space to the left and right of the expression in the numerator, putting it in display mode) are better left for the body of the problem. –  Arturo Magidin Sep 28 '11 at 18:12
    
alright thank you for the advice –  KubaSub Sep 28 '11 at 18:28
    
This is a bit roundabout, but if you are familiar with derivation formulas, you can note that this limit is the derivative of a certain function. (Edit: ... which Arturo pointed out already). Of course this isn't really using the limit laws. –  josh Sep 28 '11 at 19:36
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4 Answers

up vote 4 down vote accepted

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 $f(x) = \frac{1}{x^2}$").

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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: $\rm\:(1/f(x))'\: =\: -\: f{\:'}(x)/f(x)^2\:.$

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You may want to say "derivative rule for reciprocals", rather than "for inverses"; the latter may be confused with the Inverse Function Theorem. –  Arturo Magidin Sep 28 '11 at 18:38
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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.

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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 $h \to 0$?

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