To evaluate the limit:

$\lim_{x \to 0^+} \frac{e^x + e^{-x}}{\sin(2x)}$

Step 1: Recognize the behavior of the individual functions

• As $x \to 0^+$, both $e^x$ and $e^{-x}$ approach 1. So, $e^x + e^{-x} \to 1 + 1 = 2$.
• The function $\sin(2x)$ behaves like $2x$ when $x$ is near zero (using the small angle approximation $\sin(x) \approx x$).

Step 2: Approximate the expression

Near $x = 0$, we can approximate $\sin(2x) \approx 2x$.

Thus, the limit becomes:

$\lim_{x \to 0^+} \frac{e^x + e^{-x}}{\sin(2x)} \approx \lim_{x \to 0^+} \frac{2}{2x}$

Step 3: Simplify the expression

$\lim_{x \to 0^+} \frac{2}{2x} = \lim_{x \to 0^+} \frac{1}{x}$

As $x \to 0^+$, $\frac{1}{x} \to +\infty$.

Final Answer:

$\lim_{x \to 0^+} \frac{e^x + e^{-x}}{\sin(2x)} = \infty$

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Would you like further details on how this result was reached, or do you have any other questions? Here are some related questions:

1. What would the limit be if we were approaching $0$ from the left instead of the right?
2. How can the small angle approximation $\sin(x) \approx x$ be formally justified?
3. Can we find a similar limit involving higher powers of $x$?
4. What happens if the denominator is replaced with $\sin(x)$ instead of $\sin(2x)$?
5. How does the function $e^x + e^{-x}$ behave near other values of $x$?

Tip: Small angle approximations like $\sin(x) \approx x$ and $\cos(x) \approx 1$ are handy tools in limit evaluation, but they are valid only when $x$ is close to 0.