Limit of a function that I think is undefined

Question

The $\displaystyle\lim_{x\to \infty}\ln(x)\sin\left(\frac{1}{x}\right)$, wolfram computes it to be $1$.

However the limit of $\ln(x)$ is infinity and $\sin(1/x)$ is $0$, the product of these are undefined, am I being stupid or is this wrong?

thanks in advance.

Answer 1

Note that

$$\ln(x)\sin\left(\frac{1}{x}\right)=\frac{\ln(x)}{x}\, \frac{\sin\left(\frac{1}{x}\right)}{\frac1x}\to 0\cdot 1=0$$

thus

$$\lim_{x\to \infty}\ln(x)\sin\left(\frac{1}{x}\right)=0$$

Answer 2

Hint: L'Hopital is a coward's way to go. Somewhere you learned $0<\sin u < u$ for small positive $u.$ Thus for large $x>0,$  $0 < (\ln x)\sin(1/x) < (\ln x)/x.$