Question

$$lim_{h \to 0} \frac{e^h - 1}{2h}$$ is
(a) 0
(b) $\frac{1}{2}$
(c) 1
(d) $e$
(e) nonexistent

Explanation:

Step1: Recall the limit definition of derivative or the standard limit

We know that the standard limit is $\lim_{x \to 0} \frac{e^{x}-1}{x}=1$. In our problem, the limit is $\lim_{h \to 0} \frac{e^{h}-1}{2h}$. We can rewrite this limit as $\frac{1}{2}\lim_{h \to 0} \frac{e^{h}-1}{h}$.

Step2: Apply the standard limit

Using the standard limit $\lim_{h \to 0} \frac{e^{h}-1}{h} = 1$, we substitute this into our expression. So we have $\frac{1}{2}\times1=\frac{1}{2}$.

Answer:

B. $\frac{1}{2}$