Derivative of Exponential at Zero

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Theorem

Let $\exp x$ be the exponential of $x$ for real $x$.


Then:

$\displaystyle \lim_{x \to 0} \frac {\exp x - 1} x = 1$


Proof 1

For all $x \in \R$, we have the following:


Having verified its prerequisites, Corollary 1 to L'Hôpital's Rule yields immediately:

$\displaystyle \lim_{x \to 0} \frac {\exp x - 1} {x} = \lim_{x \to 0} \frac {\exp x} {1} = \exp 0 = 1$

$\blacksquare$


Proof 2

Note that this proof does not presuppose Derivative of Exponential Function.

We use the definition of the exponential as a limit of a sequence:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \frac{\exp h - 1} h\) \(=\) \(\displaystyle \) \(\displaystyle \frac{\lim_{n \to \infty} \left({1 + \dfrac h n}\right)^n - 1} h\) \(\displaystyle \) \(\displaystyle \)          by definition of the exponential          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \frac{\displaystyle \lim_{n \to \infty} \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\) \(\displaystyle \) \(\displaystyle \)          Binomial Theorem          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \lim_{n \to \infty} \frac{\displaystyle \sum_{k \mathop = 0}^n {n \choose k} \left({\frac h n}\right)^k - 1} h\) \(\displaystyle \) \(\displaystyle \)          as $h$ is constant          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \lim_{n \to \infty} \left({ {n \choose 0} 1 - 1 + {n \choose 1} \left({\frac h n}\right)\frac 1 h + \sum_{k \mathop = 2}^n {n \choose k} \left({\frac h n}\right)^k \frac 1 h }\right)\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \lim_{n \to \infty}1 + \lim_{n \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k-1} }{n^k}\) \(\displaystyle \) \(\displaystyle \)          Powers of Group Elements          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle 1 + h \lim_{n \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac{h^{k-2} } {n^k}\) \(\displaystyle \) \(\displaystyle \)          Powers of Group Elements          

The right summand converges to zero as $h \to 0$, and so:

$\displaystyle \lim_{h \to 0}\frac{\exp h - 1} h = 1$

$\blacksquare$


Proof 3

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \frac {e^x - 1} x\) \(=\) \(\displaystyle \) \(\displaystyle \frac {e^x - e^0} x\) \(\displaystyle \) \(\displaystyle \)          Exponential of Zero          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\to\) \(\displaystyle \) \(\displaystyle \left.{\dfrac {\mathrm d} {\mathrm dx} e^x}\right \vert_{x \mathop = 0}\) \(\displaystyle \) \(\displaystyle \)          as $x \to 0$, from definition of derivative at a point          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle e^x \vert_{x \mathop = 0}\) \(\displaystyle \) \(\displaystyle \)          Derivative of Exponential Function          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle 1\) \(\displaystyle \) \(\displaystyle \)          Exponential of Zero          

$\blacksquare$