I'm stuck on how to evaluate the following using L'Hôpital's rule:

$$\lim_{x \to \infty}\left(1 + \frac{2}{x}\right)^{3x}$$

This is a problem that I encountered on Khan Academy and I attempted to understand it using the resources there. Here are the tips given for the problem; the portion that I'm having trouble understanding is highlighted:

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I also attempted to use this video (screenshot following) to help; I understand the concepts in the video but it seems like there are some missing steps in the tips above.

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I also attempted to use WolframAlpha's step-by-step solution but it was indecipherable to me.

Any help is greatly appreciated.

  • $\begingroup$ Another technique to go over this question is here and I think it's rather simple than L'Hopital. math.stackexchange.com/questions/834589 $\endgroup$ – Saharsh Jul 23 '14 at 17:53
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    $\begingroup$ It looks as if the step that you are wondering about is $\ln(a^b)=b\ln a$, a standard property of logarithms, $\endgroup$ – André Nicolas Jul 23 '14 at 17:56
  • $\begingroup$ I'm (almost) always for using a simpler solution but the goal of the exercise where I encountered this is to learn L'Hôpital's rule so I would like to understand the solution in that context. $\endgroup$ – WXB13 Jul 23 '14 at 17:56
  • $\begingroup$ The above comment deals with the question mark you had, not with another way to find the limit. $\endgroup$ – André Nicolas Jul 23 '14 at 17:58
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    $\begingroup$ @GaryWhite: A nearly universal move when we apply the calculus to $a^b$, where $b$ is not constant, is to write $a^b=e^{b\ln a}=\exp(b\ln a)$. In our case, we have $a=1+\frac{2}{x}$ and $b=3x$. The strategy is to then find the limit of $3x\ln\left(1+\frac{2}{x}\right)$ and to use the continuity of the exponential to deduce the limit of $\left(1+\frac{2}{x}\right)^{3x}$. $\endgroup$ – André Nicolas Jul 23 '14 at 18:18

$(1+{2 \over n})^{3n} = ((1+{2 \over n})^{n})^3$.

We have $\lim_{n \to \infty} (1+ {\alpha \over n})^n = e^\alpha$.

To see the latter using l'Hôpital, let $a_n = (1+ {\alpha \over n})^n$. Then $\log a_n = n \log(1+ {\alpha \over n})= { \log(1+ {\alpha \over n}) \over {1 \over n}}$.

Note that $\lim_{n \to \infty} { \log(1+ {\alpha \over n}) \over {1 \over n}} = \lim_{x \to 0} { \log(1+ \alpha x) \over x}$. Using l'Hôpital, we see that the limit is $\alpha$, so we have $\lim_{n \to \infty} \log a_n = \alpha$,from which we get $\lim_{n \to \infty} a_n = e^\alpha$.

  • $\begingroup$ I apologize for not being specific enough. The exercise is on using L'Hôpital's rule so I'm not just interested in finding the limit but rather in taking the derivative so as to apply the rule. I've edited my post to reflect this. Thanks for your help. $\endgroup$ – WXB13 Jul 23 '14 at 17:52
  • $\begingroup$ Why the downvote? If there is something wrong I would like to fix it. $\endgroup$ – copper.hat Jul 23 '14 at 17:56
  • $\begingroup$ @GaryWhite: I have explained the reasoning above. $\endgroup$ – copper.hat Jul 23 '14 at 17:57
  • $\begingroup$ I did not downvote it. I don't even have enough of a reputation yet (15 I think) to allow me to do that. $\endgroup$ – WXB13 Jul 23 '14 at 17:57
  • $\begingroup$ @GaryWhite: My comment was directed at the downvoter... $\endgroup$ – copper.hat Jul 23 '14 at 17:58

Let $y = lim_{x→\infty}(1+\frac{2}{x} )^{3x} $

$\log y = lim_{x→\infty}(3x) ~~log~(1+\frac{2}{x} ) $

$= lim_{x→\infty} \frac{ log~(1+\frac{2}{x} )}{\frac{1}{3x} }$

$= 0/0 $ form.

Differentiate numerator and denominator and get limit value as

$\log y = lim_{x→\infty} {\frac {1} {(1+2/x)} }. (\frac {-2}{x^2}).(-\frac {x^2}{3})$


Hence, $y=e^{2/3}$


Generally when l'hospital is involved in a complex limit, you need to set the limit equal to y then apply ln to both sides. To obtain the indeterminite form you need to move a variable to the denominator. The problem usually proceeds smoothly from there. When you do solve the limit remember that it is set to ln (y), so solving for y is the final step involved.


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