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I've been reading about partial fraction expansions using L'Hopital's Theorem and have found this document.

I was wondering if it were possible to use the method mentioned in the link to find all the coefficients of a function with a perfect square as the denominator.

For example, let us suppose I had: $$ \frac{x+1}{(x+7)^{2}} = \frac{A}{(x+7)^{2}} + \frac{B}{x+7}$$

I can find $A$, the problem is, finding $B$.

This is my attempt: $$ B = \lim_{x\to-7} \frac{(x+1)(x+7)}{(x+7)^{2}} + \lim_{x\to-7} \frac{A(x+7)}{(x+7)^2} $$ $$ B = (-7+1)\lim_{x\to-7} \frac{(x+7)}{(x+7)^{2}} + A\lim_{x\to-7} \frac{(x+7)}{(x+7)^2} $$

So the problem is, the degree of the denominator is greater than that of the numerator, and if I use L'Hopital's Theorem, I can never make it so that I can sub in $-7$ without getting an undefined value.

I looked up the answer online and $ A = -6 $ and $ B = 1 $.

I should also mention, this is not at all homework, I'm just reading about partial fractions and I wondered about this.


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up vote 1 down vote accepted

There is a sign mistake (should be $B=... - ...$) and as you have to calculate $A$ first, you might as well plug it in at this point. This gives you $$ B=\frac{(x+1)(x+1-A)}{(x+7)^2} = \frac{(x+7)^2}{(x+7)^2} = 1 $$

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