Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks.

share|improve this question

1 Answer 1

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 $$ B=\frac{(x+1)(x+1-A)}{(x+7)^2} = \frac{(x+7)^2}{(x+7)^2} = 1 $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.