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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.

Evaluate the limit lim_{h -> 0} (e^{x+h} - e^x) / h.

Learn how to evaluate the limit lim_{h -> 0} (e^{x+h} - e^x) / h step by step. This solution uses properties of exponential functions, the definition of a derivative, and key calculus concepts. Ideal for students studying AP Calculus or undergraduate-level calculus.

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