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How would one find the limit of the following.

as $x\rightarrow\infty$

$(1+\frac{2}{x})^{3x}$

I did the following

$e^{\ln(1+\frac{2}{x})3x}$

$\frac{\ln(1+\frac{2}{x})}{1/3x}$

Then I did de l'Hôpital's rule.

$\frac{\frac{1}{1+2/x}\frac{-1}{x^2}}{-3/x^2}$

This is the part I having trouble in as when you try simplify the complex fraction dont you have to "flip" it.

I get

$\frac{1}{1+2/x}\frac{-2}{x^2}\frac{x^3}{-3}$

The x^2 cancel out and you are left with.

$\frac{2}{3}\frac{1}{1+2/x}$

The final limit is

$e^{\frac{2}{3}}$

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    $\begingroup$ Your error occurred in the denominator in the line right after citing l'Hôpital's rule: You wrote $-3/x^2$ instead of $-1/3x^2$. (In the line above it, the $3$ is in the denominator of the denominator, so it should have stayed there, or else been moved all the way up to the numerator of the entire expression.) $\endgroup$
    – Barry Cipra
    Oct 18, 2013 at 20:39
  • $\begingroup$ One thing is 1/3x is 3x^-1 take derivative $-3x^{-2}$ is this not -3/x^2 I guess not. But is not $\frac{-1}{3x^2}$ tje same as $\frac{1}{3}x^{-2}$ $\endgroup$
    – Fernando Martinez
    Oct 18, 2013 at 20:48
  • $\begingroup$ so the entire thing goes under the one. So if you had $\frac{1}{2}x^{-3}$ it would be $\frac{1}{1/2 x^3}$ $\endgroup$
    – Fernando Martinez
    Oct 18, 2013 at 20:54
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    $\begingroup$ $1/3x$ is $(3x)^{-1}$, or $(1/3)x^{-1}$, but not $3x^{-1}$. You have to be really careful when working with fractions. It's easy to get things mixed up. $\endgroup$
    – Barry Cipra
    Oct 18, 2013 at 21:17
  • $\begingroup$ Why not substituting $x=2y$?. It instantaneously gives $e^6$ :) $\endgroup$
    – chubakueno
    Oct 18, 2013 at 22:04

2 Answers 2

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Take $ln$ of $(1+\frac{2}{x})^{3x}$

$$L=\lim_{x\rightarrow\infty}3x\ln(1+\frac2x)$$ $$L=\lim_{x\rightarrow\infty}\frac{3\ln(1+\frac2x)}{\frac1x}$$ Now let $a=\frac1x$. If $x\rightarrow\infty$ then $a\rightarrow0^+$. So we get $$L=\lim_{a\rightarrow0^+}\frac{3\ln(1+2a)}{a}$$ Applying L'hospital $$L=3\lim_{a\rightarrow0^+}\frac{\frac{2}{1+2a}}{1}=6$$ Therefore the limit is $$e^6$$

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  • $\begingroup$ So I can take out the 3x take the 3 out as a constant $\endgroup$
    – Fernando Martinez
    Oct 18, 2013 at 20:36
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    $\begingroup$ Yes, you can do it. $\endgroup$
    – Ömer
    Oct 18, 2013 at 20:40
  • $\begingroup$ so if I had $ln(x)\frac{x}{5}$ you could take out $\frac{1}{5}$ and do $\frac{lnx}{\frac{1}{x}}$ $\endgroup$
    – Fernando Martinez
    Oct 18, 2013 at 20:42
  • $\begingroup$ Yes, that is true. $\endgroup$
    – Ömer
    Oct 18, 2013 at 20:47
  • $\begingroup$ Try and understand what he is doing: Dividing by a fraction is the same as 'multiplying by the opposite': $$\frac{a}{\frac{1}{x}}=ax.$$ You can always do this when $x$ is in the denominator, it's a great trick! $\endgroup$
    – Numbersandsoon
    Oct 18, 2013 at 20:47
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Limits of the type f(x)^g(x) where f(x) tends to 1 and g(x) to infinity i.e 1^infinity

can be done as e^((f(x)-1).g(x))

if you do that you get e^6

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  • $\begingroup$ That's way cool. Do you have a proof of this? $\endgroup$
    – Numbersandsoon
    Oct 18, 2013 at 20:45
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    $\begingroup$ I forgot, I read it somewhere and have been using it as a shortcut since, I will try it again though. $\endgroup$
    – Amandeephy
    Oct 19, 2013 at 16:44

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