I don't know what a correct answer is. Graphing the function with a computer doesn't make it too clear, but it seems to say that it approaches zero from the right. There's a good chance this problem might be expecting me to use the Squeeze Theorem, but perhaps it might be the Cauchy-Schwartz inequality. I'm not sure.
An attempt. I first look at what happens to $\pi/x$ as $x$ approaches 0 from the right. I conclude it goes to infinity. Then I consider what happens to $\sin(\pi/x)$ and I conclude it oscillates around $-1$ and $1$, that is, $$-1 \leq \sin(\pi/x) \leq 1.$$ Then I raise $e$ to both inequalities and I get $$e^{-1} \leq \sin(\pi/x) \leq e.$$ Now I multiply both inequalities by $\sqrt{x}$, getting $$\frac{\sqrt{x}}{e} \leq \sqrt{x} e^{\sin(\pi/x)} \leq e \sqrt{x}$$ and by the Squeeze Theorem I would be able to say that it goes to zero.
Related question. What happens if we let $x$ approach $0$ from the left? It seems that the function would also approach zero, but perhaps my book is avoiding the question of what happens when we consider values of the function outside of the domain — by only asking what happens when the function is not undefined.
Another possible solution. I believe the Cauchy-Schwartz inequality could also be of a solution here, but, even if it is, I wonder what an experienced eye would say the book is expecting me to do. (I have not yet used the Cauchy-Schwartz inequality in this section of the book yet.)
Reference. This exercise is number $38$ in section $2.3$ of Stewart's Calculus 6th edition translated to the Portuguese language. The original 6th edition seems to ask us to prove $\sqrt{x} [1 + \sin^2(2\pi/x)] = 0$ as $x$ approaches zero from the right. This problem from the original version seems to be concerned with the Squeeze Theorem and not the Cauchy-Schwartz inequality.
Best Answer
The $\sin$ function is bounded below by -1 and above by 1. Thus $e^{\sin\frac{\pi}{x}}$ is bounded below by $e^{-1}$ and above by $e$.Since $\sqrt x$ approaches $0$ from the right, so does your function.