# 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?

• You are not wrong, wolfram is Commented Dec 5, 2021 at 22:12
• No, Wolfram is correct. @NinadMunshi Commented Dec 5, 2021 at 22:13
• @ThomasAndrews mind explaining the little subsequence of poles? For example I would not say $\lim_{x\to\infty}x\sec^2x$ exists either for the same reason. Commented Dec 5, 2021 at 22:18
• @NinadMunshi see my answer. Functions which have undefined values don’t preclude limits as $x\to\infty,$ unless the domain 8# bounded. Commented Dec 5, 2021 at 22:23
• @ThomasAndrews you can claim whatever you want, but I haven't seen a source that defines a limit in the $x>M$ way that would allow this. I'm not saying it's not practical but it's very haphazard. Commented Dec 5, 2021 at 22:25

$$\lim_{n\to \infty} \frac{n}{2+\pm (-1)^n}$$ 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 $$f(x)\to\infty.$$

• I can't understand why from $\frac{x \ln x}{2} \to \infty$ and $\frac{x \ln x}{1 - \sin x} \ge \frac{x \ln x}{2}$ follows the final result. What theorem did you use? Commented Dec 5, 2021 at 22:33
• If $g(x)\to+\infty$ and $f(x)\geq g(x)$ for all $x>M,$ for some $M,$ then $f(x)\to+\infty.$ That is easy to prove from the definition. @rookie_of_math Commented Dec 5, 2021 at 22:41
• @rookie_of_math Refer to squeeze theorem and this one.
– user
Commented Dec 5, 2021 at 22:47
• @ThomasAndrews Thanks Commented Dec 5, 2021 at 22:58
• @user Thank you too Commented Dec 5, 2021 at 22:59

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.

• I can't understand why from $\lim_{x \to \infty} x \ln x = \infty$ and $(1)$ follows the final result. What theorem did you use? Commented Dec 5, 2021 at 22:29
• $$\lim_{x\to\infty}x\ln(x)=\infty\implies\lim_{x\to\infty}\frac{x\ln(x)}2=\infty\implies\lim_{x\to\infty}\frac{x\ln(x)}{1-\sin(x)}=\infty$$ Commented Dec 5, 2021 at 22:32
• But when $x \to \infty$, I don't have $2 = 1 - \sin x$. I have that $1 - \sin x$ oscillates between $0$ and $2$. Commented Dec 5, 2021 at 22:45
• Your answer is the same as Thomas Andrews. Am I right? So your answer uses the same theorem. Right? i.e. if $f(x) \ge g(x)$ and $g(x) \to \infty$, then $f(x) \to \infty$. Right? Commented Dec 5, 2021 at 23:04
• Yes, that is correct. And I never claimed that $2=1-\sin x$, only that $2\geqslant1-\sin x$. Commented Dec 6, 2021 at 7:26