-3
$\begingroup$

I'm trying to resolve this limit from my calculus book $$\lim_{x\to \infty} x(a+\sin(x))$$ where a is a real number. I'm asked to use the squeeze theorem. I thought that, being $\sin(x)$ a function giving values in this range: [-1,1] I could try using $(a+1)x$ and $(a-1)x$ as upper and lower bonds in the squeeze theorem(maybe splitting in cases for a>0, a=0 and a<0, if needed). My doubt is the following: does the limit exist? Because I'm confused about the existence of the limit itself. I'm imagining a function which keeps oscillating without tending to a fixed value. Could someone gently clarify this for me? ps: sorry for the terrible writing, I'm trying to use MAthjax for the first time and this isn't the result I wanted, I'll fix it asap

$\endgroup$
  • 1
    $\begingroup$ i can not read your post $\endgroup$ – Dr. Sonnhard Graubner Nov 25 '17 at 9:51
  • $\begingroup$ now it should be better $\endgroup$ – Atari96 Nov 25 '17 at 9:57
  • $\begingroup$ there is no limit $\endgroup$ – Dr. Sonnhard Graubner Nov 25 '17 at 10:00
  • $\begingroup$ try $$x_n=2n\pi$$ or $$x_n=n\frac{\pi}{2}$$ $\endgroup$ – Dr. Sonnhard Graubner Nov 25 '17 at 10:02
  • $\begingroup$ So, if I understood properly, the limit doesn't exist (indipendently from the value of the parameter a)? This makes really sense. Thank you for your help. Ok, and that's the way to show it (using succession)? Ok, clear now. Thank really. $\endgroup$ – Atari96 Nov 25 '17 at 10:03
0
$\begingroup$

The limit as $x\to\infty$ never exists in $\Bbb R$. However, if we call $A,B\in[-\infty,\infty]$, $A=\limsup\limits_{x\to\infty} x(a+\sin x)$, $B=\liminf\limits_{x\to\infty} x(a+\sin x)$, any of the following behaviours may occur, depending on the values of $a$:

  1. $B=A=+\infty$
  2. $B=0$, $A=+\infty$
  3. $B=-\infty$, $A=+\infty$
  4. $B=-\infty$, $A=0$
  5. $B=A=-\infty$

And this is indeed strongly related to the fact that $\sin x$ takes values in $[-1,1]$. I don't know exactly why the text hints towards the squeeze theorem. In a sense, you can use its "divergency case" - such as "If a function is bounded below by a function which diverges to $+\infty$, then it diverges to $+\infty$" -, but personally I can't take out of my head the notion that the author's intention was to write $x\to0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.