So I have to evaluate: $$\lim_{x\to\infty}\left(1 + \frac{2}{x}\right)^x$$

This is:

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


$$x\ln\left( 1 + \frac{2}{x} \right) = \frac{\ln\left( 1 + \frac{2}{x} \right)}{\frac{1}{x}} $$

So I should apply L'Hospital rule and calculate: $$\lim_{x\to\infty}\frac{\ln\left( 1 + \frac{2}{x} \right)}{\frac{1}{x}}$$

However I'm stuck at this step and I don't know how to move further and land at the final result ( that is, $\displaystyle{\lim_{x\to\infty}}\left(1 + \frac{2}{x}\right)^x = e^2$ )

( Alternative options are also welcome as long as they're more simple and straightforward to apply than L'Hospital rule ).

  • $\begingroup$ You can't apply L'Hopital to find the limit $\frac{1+\frac{2}{x}}{\frac{1}{x}}$ because it is not an indeterminate form. $\endgroup$ – Thomas Andrews Sep 11 '17 at 19:57
  • $\begingroup$ In general, we have $\lim_{x\to\infty}(1+k/x)^x=e^k$ where $k\in\Bbb R$. This follows easily with a few substitutions if you start with the limit definition of $e$. $\endgroup$ – Prasun Biswas Sep 11 '17 at 19:59
  • $\begingroup$ I guess I have now fixed all issues with this question. I typed it in quite fast but now it's fixed $\endgroup$ – Ruan Sep 11 '17 at 20:04
  • $\begingroup$ You've changed $1+\frac{2}{x}$ to $\ln(1+\frac{2}{x})$ but you haven't changed the rest of your argument, which would necessarily change. $\endgroup$ – Thomas Andrews Sep 11 '17 at 20:07
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    $\begingroup$ @ThomasAndrews This is what I was looking for. Write it as an answer and it will become the accepted one $\endgroup$ – Ruan Sep 11 '17 at 20:12

The key trick for computing the last limit is to set $y=\frac{1}x$ then compute the limit $\lim_{y\to 0^+} \frac{\ln(1+2y)}{y}$, which is much easier using L'Hopital.

You can apply L'Hopital because this limit is of the indeterminate form $\frac{0}{0}$.

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    $\begingroup$ This is just the definition of derivative, rather than l'Hôpital. $\endgroup$ – egreg Sep 11 '17 at 20:52
  • $\begingroup$ Well, L'Hopital is a generalization of that definition, but yes. :) @egreg $\endgroup$ – Thomas Andrews Sep 11 '17 at 20:53
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    $\begingroup$ It's a matter of principle: you use L'Hospital's theorem to compute the derivative, but you need the derivative in order to apply L'Hospital's theorem... $\endgroup$ – Alex M. Sep 13 '17 at 9:08
  • $\begingroup$ I don't think using L'Hopital here is necessarily problematic. Related discussion on meta: math.meta.stackexchange.com/q/27936/9464 $\endgroup$ – Jack Feb 24 '18 at 18:04

Since $\lim_{x\to\infty}(1 + \frac{1}{x})^x =e$, $\lim_{x\to\infty}(1 + \frac{2}{x})^x =\lim_{x\to\infty}((1 + \frac{2}{x})^{x/2})^2 =e^2 $.


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