Limit $\lim_{x\to\infty} (1+\frac{1}{x})^x$

Question

I first begin to change it into $\mathrm{e}^{x\ln\left(1+\frac{1}{x}\right)}$

I believe the way of thinking here is: A large number times a very very small number. Also, that it is approaching 1, which leads this equation to equal to e. I would just like a better explanation to understand this better! Please and thank you.

Answer 1

$$L = \lim_{x\to\infty} \left(1+\frac 1x\right)^x$$ $$\log L = \lim_{x\to\infty} x\log\left(1 + \frac 1x\right) $$$$= \lim_{x\to\infty}\frac{\log\left(1 + \frac 1x\right)}{\frac 1x} = \lim_{x\to\infty} \frac{\frac{1}{1 + \frac 1x}\cdot \frac 1{x^2}}{\frac 1{x^2}} =\lim_{x\to\infty} \frac 1{1 + \frac 1x} = 1$$ So $$\log L = 1 \implies L = e$$

Answer 2

First, the translation is $\;\mathrm e^{x\ln(1+\tfrac1x)}$.

Now $\ln(1+u)\sim_0 u$, hence $\;x\ln(1+\tfrac1x)\sim_\infty x\cdot\dfrac1x=1$, hence $\;\mathrm e^{x\ln(1+\tfrac1x)}\to \mathrm e^1=\mathrm e.$

Answer 3

$\ln(1+\dfrac1x)=\dfrac1x-\dfrac1{2x^2}+\dfrac1{3x^3}+\dots+(-1)^{k-1}\dfrac1{kx^k}+\dots$ for $|x|\gt1$.

so: $x\ln(1+\dfrac1x)=1-\dfrac1x[\dots]$

and so: $\lim_\limits{x\to\infty}x\ln(1+\dfrac1x)=1$