Is WolframAlpha wrong? or am I? Problem :
Determine a limit
$$\lim_{x\to\infty}xe^{\sin x}$$
exists or not. If it exists, find a limit.

Since $-1\le \sin x \le 1$, I can say that $\displaystyle0<\frac{1}{e}\le e^{\sin x}\le e$.
Multiply $x$ both side, $\displaystyle0<\frac{x}{e}\le xe^{\sin x}\le ex$
Since $\displaystyle\lim_{x\to\infty} \frac{x}{e}=\infty$, I think $\displaystyle\lim_{x\to\infty}xe^{\sin x}=\infty$ also.
But, WA says it is indeterminate form :

Am I correct? or Am I wrong?
If I'm wrong, where did I make a mistake?
 A: Your calculation is right.
Although the WA algorithm is almost advanced, you caught a gap there.  However, WA calculates the limit $\lim_{x\to\infty} \frac{1}{x e^ {\sin x}}=0$ correctly.  This justifies the original limit.

EDİT:
You multiplied both side of the equation $$0<\frac{1}{e}\le e^{\sin x}\le e$$ by $x$.  So, you wrote that $$0<\frac{x}{e}\le xe^{\sin x}\le ex$$
It would be a more rigorous solution if you just stated that you did this with the $x>0$ condition.
A: No, you are perfectly right.
Maybe what WolframAlpha is telling you is that it could not resolve the limit. Just because you are smarter than the computer does not mean you did something wrong.
A: Using Maple, the limit is calculated correctly ($\displaystyle\lim_{x\to\infty}xe^{\sin x}=\infty$)
But when I use "limit methods" to ask for clarification, it tries to do this:
$\displaystyle\lim_{x\to\infty}xe^{\sin x}$
$=\displaystyle(\lim_{x\to\infty}x)(\lim_{x\to\infty}e^{\sin x}) [product]$
$=\displaystyle\infty(\lim_{x\to\infty}e^{\sin x}) [identity]$
$=\displaystyle\infty(e^{\lim_{x\to\infty}{\sin x}}) [exp]$
$=\displaystyle{undefined} [sin]$
$\displaystyle\lim_{x\to\infty}xe^{\sin x}={undefined}$
Maybe, Wolfram Alpha tries to do something similar.
And the explanation provided for $\lim_{x\to\infty} \frac{1}{x e^ {\sin x}}=0$ is:
$\lim_{x\to\infty} \frac{1}{x e^ {\sin x}}$
$\displaystyle=1/{\lim_{x\to\infty} \frac{1}{x e^ {\sin x}}} [power]$
$\displaystyle=1/{(\lim_{x\to\infty}x)(\lim_{x\to\infty}e^{\sin x})} [product]$
$\displaystyle=0 [identity]$
$\lim_{x\to\infty} \frac{1}{x e^ {\sin x}} = 0$
