L'Hospital rule for $\frac{0}{0}\cdot\frac{a}{0}$

Question

Is there a solution to use the rule of L'Hospital for a boundary value problem like

$$\lim_{x\to1}\left(\dfrac{x^{2}-1}{x-1}\cdot\dfrac{a}{x^{3}-1}\right)$$

where $a> 0$ ?

I know how to solve $\dfrac{x^2-1}{x-1}$ but not how to deal with the other term.

Thank you for any help!

Answer 1

How to solve the following by L'Hopital? $$ \lim_{x\to1}\left(\frac{x^2-1}{x-1}\cdot\frac{a}{x^3-1}\right) $$

"...I know how to solve $(x^2−1)/(x−1)$ but not how to deal with the other term."

I believe here that you mean that you would understand an example like $$ \lim_{x\to1}\left(\frac{x^2-1}{x-1}\right)=\lim_{x\to1}\left(\frac{f(x)}{g(x)}\right) $$ where $f(x)=(x^2-1)$ and $g(x)=(x-1)$ and be able to solve that.

So why is the other example throwing you off? I assume that you are doing the following: $$ \lim_{x\to1}\left(\frac{x^2-1}{x-1}\cdot\frac{a}{x^3-1}\right)=\lim_{x\to1}\left(\frac{f(x)}{g(x)}\cdot\frac{a}{x^3-1}\right) $$ and are now at a complete loss as to what to do with that "third term".

The problem is simply that you chose $f(x)$ and $g(x)$ poorly! For L'Hopital to work, there can't be "another term" left over. A properly chosen $f(x)$ and $g(x)$ pair should leave nothing left over.

Try instead $f(x)=(x^2-1)\cdot a$ and $g(x)=(x-1)\cdot(x^3-1)$.

Answer 2

$\frac{0}{0} * \frac{a}{0} = \frac{0*a}{0*0}$, in which you can then use L'Hopital's rule.

In your example, we have $\frac{x^2-1}{x-1} \frac{a}{x^3-1} = \frac{a(x^2-1)}{(x-1)(x^3-1)}$. This is of the form $\frac{0}{0}$.