0
$\begingroup$

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

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

$\endgroup$
5
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $1/3x$ 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

Reset to default
3
$\begingroup$

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 $$e^6$$

$\endgroup$
6
  • $\begingroup$ So I can take out the 3x take the 3 out as a constant $\endgroup$
    – Fernando Martinez
    Oct 18, 2013 at 20:36
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ That's way cool. Do you have a proof of this? $\endgroup$
    – Numbersandsoon
    Oct 18, 2013 at 20:45
  • 1
    $\begingroup$ 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

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.