Why Wolfram-Alpha (W-A) said that $\lim_{x \to \infty} e^{\sqrt{x}}\cdot\text{sinc}(x) = 0\,$, if $\,\lim_{x \to \infty} \frac{e^{\sqrt{x}}}{x^n} = \infty$ for any $n\gg 0$??
I have tried the first one in W-A and it show that its value is zero here, even when it plot here diverges ¯_(ツ)_/¯
Even, W-A says that $\int_0^\infty e^{\sqrt{x}}\text{sinc}(x)\,dx = 3.41751$ as you can see here... It must be diverging also, isn' it??
Is counter-intuitive that a function $\neq 0$ of the form $\text{sinc}(x) =$ $\frac{\text{bounded range}\,\in \, \left[-1;\, 1\,\right]}{x}$ makes converge the function $e^{\sqrt{x}}$ when no inverse of a polynomial can $\frac{1}{x^n}$, no matter how big their exponent $n > 1$ is.
I would like to know why W-A says that is "true" for the "sinc function" (since W-A is actually really good at math): Is "truly" a "mistake"?? Or the limit is "right" and something rare is happening? If so, please explain it with the corresponding "math".
Answer to the last comment:
Gary has already given a proof that $\lim_{x \to \infty} f(x)$ for $f(x) = e^{\sqrt{x}} \frac{\sin x}{x}$ doesn't exist. I will try to explain in.
Indeed, put $x_n = 2\pi n+ \frac{\pi}2$, hence $f(x_n) = \frac{e^{\sqrt{2\pi n + \frac{\pi}2 }}}{2\pi n+ \frac{\pi}2 } \to +\infty$ as $n \to \infty$.
Put $y_n= 2\pi n$ hence $f(y_n) = 0$. Hence the Heine limit $\lim_{x \to \infty} f(x)$ doesn't exists and hence the statement from the question is not true.
If any step in solution is not clear, I'll prove it.