How to calculate $\lim_{x\to\infty}{\left(1+\frac{1}{3x}\right)}^{4x}$? 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.
 A: $$
\lim_{x\to\infty}\biggl(1+\frac1{3x}\biggr)^{4x}=\lim_{x\to\infty}\biggl(\biggl(1+\frac1{3x}\biggr)^{3x}\biggr)^{4/3}=\biggl(\lim_{x\to\infty}\biggl(1+\frac1{3x}\biggr)^{3x}\biggr)^{4/3}=e^{4/3}
$$
using the continuity of the function $x\mapsto x^{4/3}$ and the fact that
$
\lim_{x\to\infty}(1+1/x)^x=e.
$
A: 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
A: 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:
$$\left( 1+\frac 1{3x}\right)^{4x}\xrightarrow[x\to \infty]{}e^{4/3}.$$
A: 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: $\lim_{x \to \infty}\ (1+ {\frac{1}{x}})^{x}$
A: $$A= \left(1+\frac{1}{3x}\right)^{4x}\implies \log(A)=4x\log\left(1+\frac{1}{3x}\right)$$ Now, using equivalents $$\log(A)\sim 4x \times\frac{1}{3x} =\frac 43 \implies A=e^{\log(A)}\sim e^{\frac 43}$$
