0
$\begingroup$

As a homework question, a friend was asked to find:

$$\lim_{x \to \infty} \frac{e^{2x}}{x^{3}}$$

Using L'Hôpital's rule I got:

$$\begin{align} \lim_{x \to \infty} \frac{e^{2x}}{x^{3}} &= \lim_{x \to \infty} \frac{2e^{2x}}{3x^{2}} \\[12pt] &= \lim_{x \to \infty} \frac{4e^{2x}}{6x} \\[12pt] &= \lim_{x \to \infty} \frac{8e^{2x}}{6} \\[12pt] &= \lim_{x \to \infty} \frac{4e^{2x}}{3} \\[12pt] &= \infty \end{align}$$

Which agrees with WolframAlpha and also makes sense intuitively as the numerator will grow much faster than the denominator.

But the teacher said that it was indeterminate because, even after applying L'Hôpital's rule 2 times, you get $\infty/\infty$. Is this simply a result of pedagogy? When the rule is first introduced you're taught to only apply it twice (because of course not all functions will yield to L'Hôpital's rule) and then in later courses you learn more about when to stop and when to keep going? Or have I messed up?

$\endgroup$
  • 1
    $\begingroup$ Is it plus or minus infinity. $\endgroup$ – hamam_Abdallah May 8 '17 at 14:31
  • $\begingroup$ @Salahamam_Fatima I copied down the question exactly as it was, although I would venture that in this beginner's calculus class they're not worrying about that yet. However, perhaps they are and therein lies the source of my error? $\endgroup$ – Au101 May 8 '17 at 14:35
  • 1
    $\begingroup$ The teacher is wrong. You can apply L'Hopital as many times as you need. In addition, it is not the only method to calculate limits, so why should an arbitrary limitation on the number of times you can apply it (where the heck did that come from?) affect the calculation of a limit? $\endgroup$ – NickD May 8 '17 at 14:38
  • $\begingroup$ No. It is plus Infinity. As you rightly mentioned numerator grows faster than denominator $\endgroup$ – jnyan May 8 '17 at 14:38
  • $\begingroup$ you can apply l'hopital only when the limit is zero/zero or infinity/infinity - otherwise it can simply give you the wrong answer - therefore applying l'hopital a second time is liable to error $\endgroup$ – Cato May 8 '17 at 14:50
0
$\begingroup$

Did the teacher say it was indeterminate or that the limit does not exist? Some texts disagree with the language that any limit can equal infinity. Otherwise, the teacher was incorrect to say that the limit was indeterminate, and should not be telling students this.

$\endgroup$
0
$\begingroup$

Put $x=\ln (t) $ . it becomes

$$\lim_{t\to+\infty}\frac {t^2}{\ln^3 (t)} $$

but for great enough $t $ we have

the well-known inequality $$\ln (t)\leq \sqrt {t} $$ thus

$$\frac {t^2}{\ln^3 (t)}\geq \sqrt {t} $$

and the limit is $+\infty $.

$\endgroup$
  • 1
    $\begingroup$ It seems that you didn't answer the question the OP asked, did you? $\endgroup$ – Ramil May 8 '17 at 14:47
  • $\begingroup$ but it will surely help others. $\endgroup$ – hamam_Abdallah May 8 '17 at 14:48
  • $\begingroup$ The point is that he wants to be able to solve it using L'Hopital's rule, and that he is trying to learn the intricacies of the rule. There are already questions on this website talking about that inequality, but if the OP asked this, there isn't sufficient information on L'Hopital's rule - so unfortunately your post is redundant and unhelpful. $\endgroup$ – Robin Aldabanx May 8 '17 at 18:37

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.