1
$\begingroup$

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?

$\endgroup$
9
  • $\begingroup$ You are not wrong, wolfram is $\endgroup$ Commented Dec 5, 2021 at 22:12
  • 2
    $\begingroup$ No, Wolfram is correct. @NinadMunshi $\endgroup$ Commented Dec 5, 2021 at 22:13
  • $\begingroup$ @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. $\endgroup$ Commented Dec 5, 2021 at 22:18
  • $\begingroup$ @NinadMunshi see my answer. Functions which have undefined values don’t preclude limits as $x\to\infty,$ unless the domain 8# bounded. $\endgroup$ Commented Dec 5, 2021 at 22:23
  • $\begingroup$ @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. $\endgroup$ Commented Dec 5, 2021 at 22:25

2 Answers 2

3
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$ Commented Dec 5, 2021 at 22:33
  • 1
    $\begingroup$ 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 $\endgroup$ Commented Dec 5, 2021 at 22:41
  • $\begingroup$ @rookie_of_math Refer to squeeze theorem and this one. $\endgroup$
    – user
    Commented Dec 5, 2021 at 22:47
  • $\begingroup$ @ThomasAndrews Thanks $\endgroup$ Commented Dec 5, 2021 at 22:58
  • $\begingroup$ @user Thank you too $\endgroup$ Commented Dec 5, 2021 at 22:59
3
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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? $\endgroup$ Commented Dec 5, 2021 at 22:29
  • $\begingroup$ $$\lim_{x\to\infty}x\ln(x)=\infty\implies\lim_{x\to\infty}\frac{x\ln(x)}2=\infty\implies\lim_{x\to\infty}\frac{x\ln(x)}{1-\sin(x)}=\infty$$ $\endgroup$ Commented Dec 5, 2021 at 22:32
  • $\begingroup$ But when $x \to \infty$, I don't have $2 = 1 - \sin x$. I have that $1 - \sin x$ oscillates between $0$ and $2$. $\endgroup$ Commented Dec 5, 2021 at 22:45
  • $\begingroup$ 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? $\endgroup$ Commented Dec 5, 2021 at 23:04
  • $\begingroup$ Yes, that is correct. And I never claimed that $2=1-\sin x$, only that $2\geqslant1-\sin x$. $\endgroup$ Commented Dec 6, 2021 at 7:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .