4
$\begingroup$

In our readings, we had L'Hôpitals rule and defined it like that:

$\lim_{x\rightarrow x_{0}}\frac{f'(x)}{g'(x)}$

Because we had it in our readings, we are allowed to use this to find limit of functions.

Now my question is, is it possible to use this rule for products? If yes, do you think I would be allowed to do it (since we have dicussed this rule in our reading...)?

Actually, a fraction is a product at the same time, isn't it?

Because we can also write:

$\lim_{x\rightarrow x_{0}}f'(x)*\frac{1}{g'(x)}$

and it would be called product, or am I totally wrong here?

How would you use L'Hôpitals rule for products? Possible at all? I could imagine it has something to do with fraction and reciprocal. But not sure about that.

$\endgroup$
  • 2
    $\begingroup$ L'Hospital's rule is to be used (in limits) if the fraction is of the form ${0\over0}$ or ${\infty\over \infty}$. $\endgroup$ – Qwerty Jul 23 '16 at 9:49
  • $\begingroup$ I have just checked that on the internet (wikipedia) and found it can also be used for $\infty * 0$ Still, no idea how it's done and if it's allowed at all. $\endgroup$ – cnmesr Jul 23 '16 at 9:50
  • $\begingroup$ Note: Firstly, $\infty * 0$ is equivalent to ${0\over 0}$. Secondly, Why do you need it to be applicable to products? Give an example to illustrate where you felt the doubt. $\endgroup$ – Qwerty Jul 23 '16 at 9:55
  • $\begingroup$ That's the idea, and that's why I'm wondering if we can use it for products as well. Problematic task was this one: math.stackexchange.com/questions/1867695/… (Luckily, I understood the task because of the people who posted. But I'd like to know if it would have worked with Hôpital as well.) $\endgroup$ – cnmesr Jul 23 '16 at 9:57
8
$\begingroup$

That's a fairly standard rewrite.

If you need to find something like $$ \lim_{x\to a} f(x)g(x) $$ where $\lim_{x\to a}f(x)=0$ and $\lim_{x\to a}g(x)=\infty$, you can define $h(x)=\frac{1}{g(x)}$ and consider $\lim_{x\to a} \frac{f(x)}{h(x)}$. You should be able to use the form of l'Hôpital you know on that.

Of course that requires $g$ to be well-behaved around $a$.

$\endgroup$
2
$\begingroup$

You can rewrite the product as a quotient, but you have to do this before you do the derivatives.

So if you've got a limit $\lim_{x\to x_0}(f(x)g(x))$ and $f(x)\to 0$ and $g(x)\to\infty$, then you can for example decide to move $g$ to the denominator giving $f(x)/(g(x))^{-1}$. And now both numerator and denominator go to zero, thus if also the other conditons are fulfilled, you can apply l'Hôpital to get $$\lim_{x\to x_0}\frac{f(x)}{(g(x))^{-1}} = \lim_{x\to x_0}\frac{f'(x)}{-(g(x))^{-2}g'(x)} = -\lim_{x\to x_0}\frac{f'(x)}{g'(x)}(g(x))^{2}$$ Note that this is not the same as $\lim_{x\to x_0}f'(x)g'(x)$.

$\endgroup$
2
$\begingroup$

You know that you can use $\frac 0 0$ and $\frac \infty \infty$ in L'Hôpital's rule. Thus, if you have $0\cdot\infty$ where $f(k)\rightarrow0$ and $g(k)\rightarrow\infty$, you can rewrite $\lim_{k\rightarrow c}{f(k)\cdot g(k)}$ as $\lim_{k\rightarrow c}{\frac {f(k)}{\frac 1 {g(k)}}}$. Now the problem is no longer in the form $0\cdot \infty$ but $\frac 0 0$, and thus you can apply L'Hôpital's rule.

$\endgroup$
1
$\begingroup$

To answer your question you should consider what L. Hospitals rule says. I will break up the theorem to two parts: condition and conclusion. I will highlight conditions only.

Conditions: 1. $ g(x)$ and $f(x)$ should be continuously differentiable in the deleted neighborhood of the real number $a$. 2. $g'(x)$ should not be zero for all values of $x$ in the deleted neighborhood of $a$. 3. $ \displaystyle \lim _{x \rightarrow a} \frac{f(x)}{g(x)}$ should be either $\left[ 0/0 \right]$ or $ \left[ \pm \infty / \infty \right] $ indeterminate.

Now to answer your question on whether to use the theorem on products or not, I should say you must convert the products especially of the form $0 \cdot \infty $ to quotients of the form $ 0 / \frac{1}{\infty} $ or $ \infty / \frac{1}{0} $ to conform to the conditions of the L Hospitals rule.

$\endgroup$
  • $\begingroup$ Thank you it makes sense now. But as you said it seems very time consuming and actually not worth using. Anyway I was interested if it's possible at all... :P $\endgroup$ – cnmesr Jul 23 '16 at 10:35
  • 1
    $\begingroup$ @cnmesr I think that time does not matter if there is no easier way. How long it takes to solve for something should only be measured relative to how long it takes to solve it a different way, and thus this method or any other should be determined for use on a case-by-case scenario. $\endgroup$ – Simply Beautiful Art Jul 23 '16 at 13:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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