To evaluate the limit:

$\lim_{h \to 0} \frac{e^{x+h} - e^x}{h},$

we proceed as follows:

Step 1: Use the exponential property

The exponential function satisfies the identity:

$e^{x+h} = e^x \cdot e^h.$

Substitute this into the given expression:

$\frac{e^{x+h} - e^x}{h} = \frac{e^x \cdot e^h - e^x}{h}.$

Step 2: Factor out $e^x$

Factor $e^x$ from the numerator:

$\frac{e^x \cdot e^h - e^x}{h} = e^x \cdot \frac{e^h - 1}{h}.$

Since $e^x$ is independent of $h$, we can pull it outside the limit:

$\lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \cdot \lim_{h \to 0} \frac{e^h - 1}{h}.$

Step 3: Evaluate $\lim_{h \to 0} \frac{e^h - 1}{h}$

To determine this limit, recall the definition of the derivative of $e^h$ at $h = 0$:

$\frac{d}{dh}[e^h] = e^h.$

Thus, at $h = 0$, the derivative is $e^0 = 1$. By the definition of a derivative:

$\lim_{h \to 0} \frac{e^h - 1}{h} = 1.$

Step 4: Combine results

Substituting back, we get:

$\lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \cdot 1 = e^x.$

Final Answer:

$\lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x.$

Would you like a detailed explanation of any step?

Here are 5 related questions for practice:

1. Evaluate $\lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}$.
2. Show that the derivative of $e^x$ is $e^x$ using the definition of a derivative.
3. Prove that $\lim_{h \to 0} \frac{\ln(1+h)}{h} = 1$.
4. Evaluate $\lim_{h \to 0} \frac{e^{2+h} - e^2}{h}$.
5. Compute $\lim_{h \to 0} \frac{a^{x+h} - a^x}{h}$ for $a > 0$.

Tip:

For exponential functions, limits often simplify by factoring terms and using the definition of derivatives.