How does one show that $\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^x$ converges?

Question

How does one show that $\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^x$ converges? I have a book that uses this to define $e$ but they don't show that it converges. I suppose we could give it an upper and lower bound but I don't know how to do this. I've seen this question before on here but I haven't seen an answer I like yet.

Answer 1

Set $h =\frac1x\to 0~~as ~~x\to\infty$ Then, $$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x}\right)^x =\lim_{x\rightarrow\infty}\exp\left(\frac{\ln\left(1+\frac{1}{x}\right)}{\frac1x}\right) =\lim_{h\rightarrow 0}\exp\left(\frac{\ln\left(1+h\right)}{h}\right) = e$$

Given that $$\lim_{h\rightarrow 0}\frac{\ln\left(1+h\right)}{h} = 1$$

Edit: For OP doubt. I think he to know how can want guess the existence of $e?$ without appealling $\ln$ and $\exp$.

In fact Euler Predicted the existence of $e$ before the creation of the functions $\ln$ and $\exp$: it is explained here: https://www.youtube.com/watch?v=AuA2EAgAegE

Answer 2

Consider $$A=\left(1+\frac{1}{x}\right)^x\implies \log(A)=x\log\left(1+\frac{1}{x}\right)$$ and use equivalents $$\log\left(1+\frac{1}{x}\right)\sim \frac{1}{x}$$ making $$\log(A)\sim 1\implies A \sim e$$

Answer 3

Consider $\ln b = \int\limits_1^b\frac{dx}{x}$ with $b = 1+\frac{1}{n}$ then

$\ln(1+\frac{1}{n}) = \int\limits_1^{1+\frac{1}{n}}\frac{dx}{x}$

The area represented by this integral lies between the area of two rectangles of height $\frac{n}{n+1}$ and $1$ and base $\frac{1}{n}$. These rectangles have areas $\frac{1}{1+n}$ and $\frac{1}{n}$, therefore:

$\frac{1}{n+1}\leq\ln\left(1+\frac{1}{n}\right)\leq\frac{1}{n}$

Multiplying through by $n$ and using $n\ln a = \ln a^n$:

$\frac{n}{n+1}\leq\ln\left((1+\frac{1}{n})^n\right)\leq1$

Since $\lim\limits_{n\to\infty}\frac{n}{n+1} = 1$ the middle quantity must approach $1$ by the squeeze theorem:

$\lim\limits_{n\to\infty}\ln\left((1+\frac{1}{n})^n\right)=1$

Now apply the continuous function $e^x$ to obtain the desired result:

$e^1 = e^{\lim\limits_{n\to\infty}\ln\left((1+\frac{1}{n})^n\right)} = \lim\limits_{n\to\infty}e^{\ln\left((1+\frac{1}{n})^n\right)} = \lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n$

Check out Rogawski Calculus section 7.5 pp 374-375.