2
$\begingroup$

I got a lot of limits questions that I am able to find there limits, but I do not know if they meet the qualification to be l'Hopital's or not. SO how to know that ?

For example:

Q1) $ \lim_{t \rightarrow \infty} \dfrac{(\ln t )^ 2}{t}$

I would say yes, infinity/infinity

Q2) $ \lim_{y \rightarrow 0} \dfrac{2y}{y^2}$

I would say yes 0/0

Q3) $ \lim_{x \rightarrow \infty} \dfrac{e^{−x}}{ 1 + \ln x }$

I would say no, 0/number

Q4) $\lim_{\theta \rightarrow 0} \dfrac{\arctan \theta}{7\theta}$

I would say yes, 0/0

Q5) $\lim_{x \rightarrow 0+ } \dfrac{\cot x}{\ln x}$

I don't know if the 0+ would make a difference of the 0. But The answer would be no because 0/undefined

So bottom line is there any rules to know if it is l'Hospital's or not? I just feel that 0/0 and infinity/infinity are the ones that can determine. BUt is there any other forms ? such as 0/1 or 1/0 or infinity/0 ?

$\endgroup$
  • $\begingroup$ $Q3$ is $\frac{0}{\infty}$, but still no. $Q5$ is $\frac{\infty}{-\infty}$, so yes. $\endgroup$ – Joe Johnson 126 Jul 8 '14 at 1:27
  • $\begingroup$ There are also other forms, like $\infty^0$, etc. You can find them here: en.wikipedia.org/wiki/Indeterminate_form $\endgroup$ – Joe Johnson 126 Jul 8 '14 at 1:29
  • $\begingroup$ @JoeJohnson126 how come that Q5 is ended to be ∞/−∞ if the lim x--->0 $\endgroup$ – John Jul 8 '14 at 1:37
  • $\begingroup$ @JoeJohnson126: The limit of cotangent at zero from the right ("$0^+$") is $+\infty$. The limit of $\ln x$ at zero from the right is $-\infty$. Have you graphed these functions recently so that their behaviour at zero is firmly in your memory? $\endgroup$ – Eric Towers Jul 8 '14 at 1:39
  • $\begingroup$ @EricTowers Yes. That's why I typed $\frac{\infty}{-\infty}$. I don't see what the problem is. $\endgroup$ – Joe Johnson 126 Jul 8 '14 at 1:48
4
$\begingroup$

The only allowable forms by L'Hopital are $$\frac{0}{0} \,\,\text{and} \,\, \frac{\pm \infty}{\pm \infty} \,\,.$$ Hopefully this suffices to answer your question. If not, let me know. Note the $\pm$ in both the numerator and denominator of the latter form. Usually, the aim is to play with the limit until it is in one of these forms and then apply L'Hopital's rule. Summarizing, there are no other forms (even indeterminate forms) that are allowable under this rule.

$\endgroup$
  • $\begingroup$ Why not? Every question should have an accepted answer. @CameronWilliams $\endgroup$ – afedder Jul 8 '14 at 2:07
  • $\begingroup$ What do you mean "if that"? $\endgroup$ – afedder Jul 8 '14 at 2:08
  • $\begingroup$ Disregard that. I totally missed the very last line. $\endgroup$ – Cameron Williams Jul 8 '14 at 2:22
  • $\begingroup$ You can use L'Hopital's rule as long as the denominator is infinite. $\endgroup$ – Brad Jul 8 '14 at 2:24
  • 1
    $\begingroup$ @Brad: didn't you contradict this statement in your answer? $\lim_{x \to \infty} \dfrac{x+\sin(x)}{x}$ is a case where "the denominator is infinite" (i.e. has an infinite limit) and l'Hopital's rule can't be used. $\endgroup$ – Robert Israel Jul 8 '14 at 4:42
2
$\begingroup$

You can use L'Hopital when

$$\lim_{x\to a} \dfrac{f'(x)}{g'(x)} = L \in [-\infty,\infty]$$

and if either

$$\lim_{x\to a} f(x) = \lim_{x\to a} g(x) = 0 \quad \text{or}\quad \lim_{x\to a} g(x) = \pm\infty$$

Each of these conditions are important. For example, you cannot use L'Hopitals rule to determine $\lim_{x\to \infty}\frac{x+\sin(x)}{x}$ even though the denominator tends to $\infty$. This is because the limit $\lim_{x\to \infty}\frac{1+\cos(x)}{1}$ does not exist.

To answer your question, there are no other forms which L'Hopital's rule may be applied. You may be able to convert your limit into a form where the rule can be applied but you cannot directly apply it to other limits.

$\endgroup$

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.