To evaluate the limit

$\lim_{x \to 0} \frac{3^x - 3^{-x}}{3^x + 3^{-x}},$

we'll simplify and compute it step by step.

Step 1: Express in terms of exponentials

Recall that $3^x = e^{x \ln 3}$ and $3^{-x} = e^{-x \ln 3}$. So, the expression becomes:

$\frac{e^{x \ln 3} - e^{-x \ln 3}}{e^{x \ln 3} + e^{-x \ln 3}}.$

This expression looks similar to the hyperbolic tangent function, so we can try to rewrite it using hyperbolic functions.

Step 2: Recognize the hyperbolic tangent form

We recall the identities for hyperbolic sine and cosine:

$\sinh(a) = \frac{e^a - e^{-a}}{2}, \quad \cosh(a) = \frac{e^a + e^{-a}}{2}.$

Thus, the expression simplifies to:

$\frac{2 \sinh(x \ln 3)}{2 \cosh(x \ln 3)} = \tanh(x \ln 3).$

Step 3: Take the limit as $x \to 0$

We know that $\tanh(y) \to 0$ as $y \to 0$. Here, $y = x \ln 3$, so as $x \to 0$, $x \ln 3 \to 0$.

Thus,

$\lim_{x \to 0} \tanh(x \ln 3) = 0.$

Final Answer:

$\lim_{x \to 0} \frac{3^x - 3^{-x}}{3^x + 3^{-x}} = 0.$

Would you like more details on any of the steps? Here are some related questions to explore:

1. What is the difference between hyperbolic functions and trigonometric functions?
2. How do you use L'Hôpital's Rule to solve limits involving exponentials?
3. Can limits involving hyperbolic functions be simplified in different ways?
4. What are the properties of the hyperbolic tangent function?
5. How does the limit of an exponential function behave as $x \to 0$?

Tip: When dealing with exponentials and limits, rewriting in terms of hyperbolic or trigonometric functions often simplifies the process!