How can we prove $\lim_ \limits{x\to0}(1+x)^{\frac{1}{x}}=e$ without using La Hôpital's Rule?

Question

The limit $\lim\limits_{x\to0}(1+x)^{\frac{1}{x}}=e$ can be proved using La Hôpital's Rule but is it possible to prove this limit using the definition of the limit?

Answer 1

Set $y=\frac{1}{x}$, and thus $$\lim_{x\to 0^+}(1+x)^{1/x}=\lim_{y\to +\infty }\left(1+\frac{1}{y}\right)^y.$$

Answer 2

Assume the limit exists and the value is some $l\in\mathbb{R}$, that is $$\lim\limits_{x\to0}(1+x)^{\frac{1}{x}}=l.$$ Consider the logarithm in both sides of the expression, $$\log\left(\lim\limits_{x\to0}(1+x)^{\frac{1}{x}}\right)=\log l.$$ This is equivalent to $$\lim\limits_{x\to0}{\frac{1}{x}}\log(1+x)=\log l.$$ Now, since $\log(1) = 0$, we can write ${\frac{1}{x}}\log(1+x)$ as $${\frac{1}{x}}\log(1+x) = {\frac{\log(1+x)-\log(1)}{x}},$$ and then, by the definition of the derivative, the limit on the left hand side is equal to $(log(x))'|_{x=1} = 1$. Thus, $$1=\log l,$$ which means that the value $l$ is $e$.

Answer 3

Consider $$A=(1+x)^{\frac{1}{x}}\implies \log(A)=\frac{1}{x}\log(1+x)$$ Now, remembering equivalents for small $x$ $$\log(1+x)\sim x$$ or Taylor series $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ you arrive to $$\log(A)\sim 1$$ or $$\log(A) =1-\frac{x}{2}+O\left(x^2\right)$$

Answer 4

If you don't define $e$ as that limit, you can do what follows:

$$\left( 1 + \frac1{x}\right)^x = \exp\left( x \ln \left(1 + \frac1{x}\right) \right) = \exp\left( x \cdot \left( \frac1{x} + O\left(\frac1{x} \right) \right) \right) = e \cdot \exp( O(1)) \to e$$