Evaluate ${\lim }_{x\to 0}\frac{e^x-1}{x}$.

Solution

1. Recognize the definition of the derivative of $f(x)=e^x$ at $x=0$.
2. The derivative is $f^{\prime }(x)=e^x$.
3. At $x=0$, $$\begin{array}{rl}{f}^{\prime }(0)& =e^0\\ & =1\end{array}$$.
4. The limit ${\lim }_{x\to 0}\frac{e^x-e^0}{x-0}$ is exactly $f^{\prime }(0)$.
5. Hence the answer is (B).