To solve the limit

$\lim_{x \to 0} (1 + 2x)^{\csc x},$

Step 1: Rewrite using logarithms

Let $y = (1 + 2x)^{\csc x}$. Taking the natural logarithm of both sides: $\ln y = \csc x \cdot \ln(1 + 2x).$ We know $\csc x = \frac{1}{\sin x}$, so this becomes: $\ln y = \frac{\ln(1 + 2x)}{\sin x}.$

Step 2: Expand $\ln(1 + 2x)$ for small $x$

Using the Taylor expansion of $\ln(1 + u)$ near $u = 0$, we have: \ln(1 + 2x) \approx 2x \quad \text{(for small x)}.

Step 3: Expand $\sin x$ for small $x$

Using the Taylor expansion of $\sin x$ near $x = 0$, we have: \sin x \approx x \quad \text{(for small x)}.

Step 4: Simplify the expression

Substitute these approximations into the logarithmic form: $\ln y \approx \frac{2x}{x} = 2.$

Thus, $\ln y \to 2$ as $x \to 0$, which implies: $y \to e^2.$

Final Answer:

$\lim_{x \to 0} (1 + 2x)^{\csc x} = e^2.$

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Would you like a detailed explanation of any step or have further questions? Here are some related questions to expand your understanding:

1. How does the Taylor expansion simplify logarithmic and trigonometric functions for small values of $x$?
2. Can this limit be solved using L’Hôpital’s Rule instead of logarithmic techniques?
3. How does the behavior of $(1 + kx)^n$ change when $n$ is a function of $x$?
4. What are some common pitfalls when solving exponential limits like this?
5. How does this problem illustrate the connection between exponential and logarithmic functions?

Tip: Always check the domain of the functions in a limit problem to ensure they are well-defined. For example, $\csc x$ is undefined at $x = 0$, so we analyze limits around this point.