To solve $\lim_{x \to 0^+} \frac{x}{e^{(x-1) \csc x}}$ step by step, we proceed as follows:

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Step 1: Rewrite the expression

The given limit is: $\lim_{x \to 0^+} \frac{x}{e^{(x-1) \csc x}}$

Here, $\csc x = \frac{1}{\sin x}$. Substituting this, we rewrite the expression as: $\lim_{x \to 0^+} \frac{x}{e^{(x-1) \cdot \frac{1}{\sin x}}}$

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Step 2: Simplify the exponent

The exponent of $e$ is: $(x-1) \csc x = (x-1) \cdot \frac{1}{\sin x}$

Split the expression into two terms: $(x-1) \cdot \frac{1}{\sin x} = \frac{x}{\sin x} - \frac{1}{\sin x}$

Thus, the expression becomes: $\frac{x}{e^{\frac{x}{\sin x} - \frac{1}{\sin x}}}$

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Step 3: Analyze $\frac{x}{\sin x}$ as $x \to 0^+$

As $x \to 0^+$, $\frac{x}{\sin x} \to 1$ since $\sin x \sim x$ for small $x$.

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Step 4: Analyze $-\frac{1}{\sin x}$ as $x \to 0^+$

As $x \to 0^+$, $\sin x \sim x \implies \frac{1}{\sin x} \to \frac{1}{x}$

Thus, the exponent becomes: $\frac{x}{\sin x} - \frac{1}{\sin x} \to 1 - \frac{1}{x}$

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Step 5: Simplify the limit expression

The expression becomes: $\lim_{x \to 0^+} \frac{x}{e^{1 - \frac{1}{x}}}$

Here, $e^{1 - \frac{1}{x}}$ grows very large as $x \to 0^+$, since $\frac{1}{x} \to \infty$. Specifically: $e^{1 - \frac{1}{x}} \sim e^{-\frac{1}{x}}$

Thus, the fraction simplifies to: $\frac{x}{e^{1 - \frac{1}{x}}} \sim \frac{x}{e^{-\frac{1}{x}}} = x e^{\frac{1}{x}}$

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Step 6: Analyze $x e^{\frac{1}{x}}$ as $x \to 0^+$

As $x \to 0^+$, $e^{\frac{1}{x}}$ grows much faster than any polynomial or $x$, leading to: $x e^{\frac{1}{x}} \to 0$

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Final Answer:

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

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Would you like further clarifications?

Here are some related questions to expand your understanding:

1. How does the substitution $\csc x = \frac{1}{\sin x}$ simplify the calculation?
2. Why does $\frac{x}{\sin x} \to 1$ as $x \to 0$?
3. What is the role of exponential growth ($e^{-\frac{1}{x}}$) in this problem?
4. How would this limit change if the expression in the denominator were $e^{x \csc x}$?
5. Can L'Hôpital's Rule be used to verify this result?

Tip: For limits involving $\sin x$ and $\csc x$, it is often helpful to approximate $\sin x \sim x$ for small $x$.