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I've been banging my head all day long over this question, would apprecate some help :)

$$f(x) = \frac{1}{x^2-1} - \frac{a}{x^3-1},\qquad a\in\mathbb{R}.$$ For which values of $a$ does $\lim\limits_{x \to 1} f( x)$ exist?

Also it was requested to use only the $\epsilon$-$\delta$ definition.

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Hint: Note that $x^3 - 1 = (x-1)(x^2 + x + 1)$, and then subtract the two fractions by finding the least common denominator. –  JavaMan Jan 13 '12 at 17:21

1 Answer 1

Both denominators are $0$ when $x=1$; therefore $(x-1)$ is a factor of both. So factor them, then find the common denominator: $$ \begin{align} & \frac{1}{x^2-1} - \frac{a}{x^3-1} = \frac{1}{(x-1)(x+1)} - \frac{a}{(x-1)(x^2+x+1)} \\ \\ & = \frac{x^2+x+1}{(x-1)(x+1)(x^2+x+1)} - \frac{a(x+1)}{(x-1)(x+1)(x^2+x+1)} \end{align} $$ The limit can exist only if the numerator $$ x^2+x+1 - a(x+1) $$ is $0$ when $x=1$. And it is equal to $3-2a$ when $x=1$.

Can you take it from there?

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yeah I understand the idea, thanks for the help :) –  sony jimbo Jan 13 '12 at 17:31

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