The Number e as a Limit

Problem 157: Show that the number $e$ can be calculated as the limit

(1) $\begin{equation*} e = \lim_{x \to 0}(1+x)^{1/x}.\end{equation*}$

Solution: Let $f(x) = \ln(x)$ . Then $f'(x) = \frac{1}{x} = 1$ at $x=1$ . But using the definition of derivative,

(2) $\begin{align*} f'(x) & = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ f'(1) & = \lim_{h \to 0} \frac{f(1+h)-f(1)}{h} \\ & = \lim_{x \to 0} \frac{f(1+x)-f(1)}{x} \\ & = \lim_{x \to 0} \frac{\ln(1+x)-\ln(1)}{x} \\ & = \lim_{x \to 0} \frac{\ln(1+x)}{x} \\ & = \lim_{x \to 0} \ln(1+x)^{1/x} \\ & = \ln \left( \lim_{x \to 0} (1+x)^{1/x} \right) \quad \text{(ln is continuous)}. \end{align*}$

Because $f'(1)=1$ and $f'(1)$ is given by (2), they must be equal. That is,

(3) $\begin{align*} 1 & = \ln \left( \lim_{x \to 0} (1+x)^{1/x} \right) \\ e^1 & = e^{ \ln \left( \lim_{x \to 0} (1+x)^{1/x} \right) } \\ e & = \lim_{x \to 0}(1+x)^{1/x}. \end{align*}$

That is,

(4) $\begin{equation*} \boxed{e & = \lim_{x \to 0}(1+x)^{1/x}}.\end{equation*}$

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