To prove that

$\lim_{x \to 0^+} \sqrt{x} \, e^{\sin(\pi/x)} = 0,$

we need to analyze the behavior of each component of the expression as $x \to 0^+$.

Step-by-Step Solution

1. Analyze $\sqrt{x}$ as $x \to 0^+$:

   As $x \to 0^+$, we see that $\sqrt{x} \to 0$. This suggests that $\sqrt{x}$ will tend to zero, which might make the entire expression go to zero. However, we must also consider the behavior of the exponential term $e^{\sin(\pi/x)}$.

2. Analyze $e^{\sin(\pi/x)}$ as $x \to 0^+$:

   As $x \to 0^+$, the term $\sin(\pi/x)$ oscillates between $-1$ and $1$ because $\pi/x$ grows without bound and causes the sine function to oscillate rapidly. Therefore, $-1 \leq \sin(\pi/x) \leq 1.$ This gives bounds on $e^{\sin(\pi/x)}$: $e^{-1} \leq e^{\sin(\pi/x)} \leq e^1.$ Thus, $e^{\sin(\pi/x)}$ is bounded as $x \to 0^+$, staying within the range $[e^{-1}, e]$.

3. Combine the Results:

   Now we can combine these observations: $\sqrt{x} \, e^{\sin(\pi/x)}.$ Since $\sqrt{x} \to 0$ as $x \to 0^+$ and $e^{\sin(\pi/x)}$ is bounded (it does not diverge as $x \to 0^+$), the product $\sqrt{x} \, e^{\sin(\pi/x)}$ will also tend to zero.

4. Conclusion:

   By the Squeeze Theorem, we conclude that $\lim_{x \to 0^+} \sqrt{x} \, e^{\sin(\pi/x)} = 0.$

Final Answer

$\lim_{x \to 0^+} \sqrt{x} \, e^{\sin(\pi/x)} = 0.$

Would you like a deeper explanation on any part of this solution, or do you have any questions?

Here are some related questions for further exploration:

1. How does the Squeeze Theorem work, and when is it used in limit problems?
2. What happens if we replace $\sin(\pi/x)$ with a different oscillatory function, like $\cos(\pi/x)$?
3. How would the limit change if $\sqrt{x}$ were replaced by $x^n$ for some power $n$?
4. What are some examples of limits where the exponential function plays a critical role?
5. How do bounded functions interact with terms that approach zero in limits?

Tip: In limit problems, if one term approaches zero and the other is bounded, the entire expression often tends to zero.