To solve the limit

$\lim_{x \to +\infty} \left(\frac{\sin\left(e^{-x}\right)}{e^{-x}}\right),$

Step-by-Step Solution:

1. Examine the behavior of $e^{-x}$:

   • As $x \to +\infty$, $e^{-x} \to 0^+$. Therefore, the argument of the sine function $e^{-x}$ also approaches 0.

2. Simplify the fraction:

   • Rewrite the expression: $\frac{\sin\left(e^{-x}\right)}{e^{-x}}.$ Let $u = e^{-x}$. Then, as $x \to +\infty$, $u \to 0^+$. The limit becomes: $\lim_{u \to 0^+} \frac{\sin(u)}{u}.$

3. Apply the standard limit result:

   • It is a well-known limit that: $\lim_{u \to 0} \frac{\sin(u)}{u} = 1.$

4. Conclude the result:

   • Hence, the given limit evaluates to: $\lim_{x \to +\infty} \left(\frac{\sin\left(e^{-x}\right)}{e^{-x}}\right) = 1.$

Final Answer:

$\lim_{x \to +\infty} \left(\frac{\sin\left(e^{-x}\right)}{e^{-x}}\right) = 1.$

Let me know if you'd like more details or have questions about this solution.

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Relative Questions:

1. Why does $\frac{\sin(u)}{u}$ approach 1 as $u \to 0$?
2. What is the significance of the substitution $u = e^{-x}$?
3. How does the behavior of $e^{-x}$ as $x \to \infty$ influence the limit?
4. Are there cases where $\lim_{u \to 0} \frac{\sin(u)}{u}$ might not hold?
5. How would the limit change if $\sin(e^{-x})$ was replaced with another function like $\cos(e^{-x})$?

Tip:

Understanding standard limits like $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ is crucial, as they often simplify complex problems!