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The limit $$\lim_{x\to\infty}{\displaystyle \left(1+\frac{1}{3x}\right)}^{4x}$$ apparently has a value of $e^{4/3}$. I can't see why this would be the case.

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  • $\begingroup$ @GuyFsone: the accepted answer to such question is frankly horrible, and in order to prove $\lim_{x\to +\infty}\left(1+\frac{1}{3x}\right)^{4x}=e^{4/3}$ from $\lim_{x\to +\infty}\left(1+\frac{1}{x}\right)^{x}=e$ one still needs a small extra, like a suitable substitution. $\endgroup$ Nov 9 '17 at 18:27
  • $\begingroup$ The substitution is $X=3x$. Is that too difficult. $\endgroup$
    – Guy Fsone
    Nov 9 '17 at 18:30
  • $\begingroup$ @GuyFsone: I agree, but still, not enough ground to mark this question as a duplicate. $\endgroup$ Nov 9 '17 at 18:36
  • $\begingroup$ math.stackexchange.com/questions/295584/… for you:) $\endgroup$
    – Guy Fsone
    Nov 10 '17 at 16:16
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You're sure familiar with this well-know fact (substituting $r$ with $1$ will give you $e$ and this actually is the definition of the number $e$ that's commonly used in elementary calculus):

$$ \lim_{n\to\infty}\left(1+\frac{r}{n}\right)^{n}=e^{r} $$

So, here's what you get (it's pretty clear that if $x$ goes to infinity, so does the quantity $4x$):

$$ \lim_{x\to\infty}\left(1+\frac{1}{3x}\right)^{4x}= \lim_{x\to\infty}\left(1+\frac{1}{3x}\cdot\frac{4}{4}\right)^{4x}= \lim_{4x\to\infty}\left(1+\frac{4/3}{4x}\right)^{4x}=e^{4/3} $$

The answer that you posted is apparently correct. So, what exactly is your problem?

Useful video materials:
https://www.youtube.com/watch?v=8HpvEANFQ7Q
https://www.youtube.com/watch?v=n0VFDT3ZObY

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  • $\begingroup$ I wasn't aware of the definition of $e^r$ that you mentioned in your answer, and so will take it as a given. $\endgroup$
    – user98937
    Nov 9 '17 at 13:45
  • $\begingroup$ I only knew the answer, but did not know how to arrive at it. $\endgroup$
    – user98937
    Nov 9 '17 at 13:54
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    $\begingroup$ $\lim_n\left(1+\dfrac 1n\right)^n$ is a definition of $e$, but certainly not the definition. Perhaps the most common definition is that $e = \operatorname{exp}(1)$ where $\operatorname{exp}$ is the unique solution to the differential equation $y' = y$ with $y(0) = 1$ $\endgroup$ Nov 9 '17 at 17:43
  • $\begingroup$ See my math.stackexchange.com/a/3212948/403337. I believe I read this proof in Best and Penner. $\endgroup$
    – user403337
    May 11 '19 at 13:39
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$$ \lim_{x\to\infty}\biggl(1+\frac1{3x}\biggr)^{4x}=\lim_{x\to\infty}\biggl(\biggl(1+\frac1{3x}\biggr)^{3x}\biggr)^{4/3}=\biggl(\lim_{x\to\infty}\biggl(1+\frac1{3x}\biggr)^{3x}\biggr)^{4/3}=e^{4/3} $$ using the continuity of the function $x\mapsto x^{4/3}$ and the fact that $ \lim_{x\to\infty}(1+1/x)^x=e. $

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You can write it like this:

$$\left( 1+\frac 1{3x}\right)^{4x}=e^{4x\log(1+1/3x)}.$$

Then if you do the change of variable $y=3x$, you get:

$$e^{4/3y\log(1+1/y)}.$$

It is a known limit (to prove it, you can use the fact that $\log(1+x)\sim x$ is $x$ goes to $1$) that:

$$y\log(1+1/y)\xrightarrow[y\to \infty]{}1$$

so finally:

$$\left( 1+\frac 1{3x}\right)^{4x}\xrightarrow[x\to \infty]{}e^{4/3}.$$

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    $\begingroup$ All is fine...... $\endgroup$
    – Green.H
    Nov 9 '17 at 13:38
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Setting $u=3x$ we get

$$ \lim_{x\to\infty}{ \left(1+\frac{1}{3x}\right)}^{4x} =\lim_{u\to\infty} \left[\left(1+\frac{1}{u}\right)^{u}\right]^{4/3} = e^{4/3}$$

Given that, $$ \lim_{u\to\infty} \left(1+\frac{1}{u}\right)^{u}=e $$

A limit to infinity: $\lim_{x \to \infty}\ (1+ {\frac{1}{x}})^{x}$

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$$A= \left(1+\frac{1}{3x}\right)^{4x}\implies \log(A)=4x\log\left(1+\frac{1}{3x}\right)$$ Now, using equivalents $$\log(A)\sim 4x \times\frac{1}{3x} =\frac 43 \implies A=e^{\log(A)}\sim e^{\frac 43}$$

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