Limit calculation: $\lim_{x \to +\infty}\frac{x \ln x}{1 - \sin x}$

I have to calculate the following limit

$$\lim_{x \to +\infty} \frac{x \ln x}{1 - \sin x}$$

According Wolfram Alpha the limit exists and is $+\infty$ but I think it does not exist since $\lim_{x \to +\infty} 1 - \sin x$ does not exist. So: why am I wrong?

Solution 1:

$$\lim_{n\to \infty} \frac{n}{2+\pm (-1)^n}$$ is infinity even though the denominator does not converge.

The real problem with your function is that it isn’t defined for all of $\mathbb R.$ But if $f$ is a function defined on a subset of $\mathbb R$ with no real upper bound, we can still define $\lim_{x\to\infty} f(x).$

In your case, you easily get, where $f$ is defined, and $x>1,$ $$f(x)=\frac{x\ln x}{1-\sin x}\geq \frac{x\ln x}{2},$$ since $0<1-\sin x\leq 2$ in the domain of $f.$

Since $\frac{x\ln x}2\to\infty,$ this means $f(x)\to\infty.$

Solution 2:

The quotient $\dfrac{x\ln(x)}{1-\sin(x)}$ is undefined when $x\in\dfrac\pi2+2\pi\Bbb Z$. Otherwise, $1-\sin(x)\leqslant2$, and therefore$$\frac{x\ln(x)}{1-\sin(x)}\geqslant\frac{x\ln(x)}2.\tag1$$Since$$\lim_{x\to\infty}x\ln(x)=\infty,$$it follows from $(1)$ that$$\lim_{x\to\infty}\frac{x\ln(x)}{1-\sin(x)}=\infty$$too.