23

Q. If $f(x)=\lim\limits_{n\to\infty}\dfrac{\log(2+x)-x^{2n}\sin(x)}{1+x^{2n}}$, then explain why the function does not vanish anywhere in the interval $[0,\pi/2]$, although $f(0)$ and $f(\pi/2)$ differ in sign.

My solution:

First, we simplify the limit a bit.

$$f(x)=\lim_{n\to\infty}\frac{\log(2+x)-x^{2n}\sin(x)}{1+x^{2n}}=\lim_{n\to\infty}\frac{\log(2+x)-(x^{2n}+1-1)\sin(x)}{1+x^{2n}}\\ \implies f(x)=\lim_{n\to\infty}\frac{\log(2+x)+\sin(x)}{1+x^{2n}}-\sin(x)$$

Now, we divide the interval $[0,\pi/2]$ into three parts, viz., $x\in [0,1)$, $x=1$ and $x\in (1,\pi/2]$.

For the first part, i.e., when $x\in [0,1)$, we have $|x|\lt 1$ and hence $1+x^{2n}\to 1+0=1$ as $n\to\infty$, hence, by the quotient rule of limits, it reduces to,

$$f(x)=\log(2+x)+\sin(x)-\sin(x)=\log(2+x)~\forall~x\in [0,1)$$

At $x=1$, we have $f(x)=\lim\limits_{n\to\infty}\frac{\log3+\sin(1)}{1+1^{2n}}-\sin(1)=\frac 12(\log3-\sin1)$

For $x\in (1,\pi/2]$, we have $|x|\gt 1$, hence $1+x^{2n}\to\infty$ as $n\to\infty$. But $\log(2+x)$ and $\sin(x)$ are finite constants for a particular $x$. Hence, the limit goes to $0$ as $n\to\infty$ and we get $f(x)=0-\sin(x)=-\sin(x)~\forall~x\in (1,\pi/2]$.

So, we write our results collectively as,

$$f(x)=\begin{cases}\begin{align}\log(2+x)&&\forall~x\in [0,1)\\ \frac 12(\log3-\sin1)&&\textrm{at }x=1\\ -\sin(x)&&\forall~x\in (1,\pi/2]\end{align}\end{cases}$$

It's easy to verify now that $f(x)$ doesn't vanish in $[0,\pi/2]$ since, for $x\in [0,1)$, we have $f(x)=\log(2+x)\gt \log1=0$ since logarithm is strictly increasing. At $x=1$, the function value is obviously not $0$ since $\log3\neq\sin1$ and for $x\in (1,\pi/2]$, we have $f(x)=-\sin(x)\in [-1,-\sin 1)$ since sine is strictly increasing in $[0,\pi/2]$.

Now, the explanation behind why $f(x)$ doesn't vanish even when $f(0)$ and $f(\pi/2)$ differ in sign would be that $f(x)$ isn't continuous on $[0,\pi/2]$, more specifically it's discontinuous at $x=1$ with left hand limit being $\log3$ and right hand limit being $-\sin1$ and hence Bolzano's theorem isn't applicable for $f(x)$ on $[0,\pi/2]$.

Comments about my solution and improvements are welcome. Thanks! :)

learner
  • 1,407
  • 3
    Nothing to add or rectify. +1. – Did Nov 25 '15 at 10:27
  • @Did, ah, glad to hear that. My batchmates were having a tough time solving this and I gave it a try but wasn't sure if my solution was rigorous enough (or even correct for that matter) to write in the exam (this question is actually from a 2011 B.Sc 1st year paper). I feel much relieved now! ^_^ – learner Nov 25 '15 at 10:34
  • 1
    The convergence is not uniform so the limit function need not be continuous. A simple example of this effect is $f_n(x)=x^n$ for $x\in [0,1]$. A graph of $f_n$ in your Q for some $n$ might give you a good idea of what is going on. Well done. – DanielWainfleet Nov 25 '15 at 17:02
  • Well done. Just comment here so I can come back later :). – Hoang-Ngan Nguyen Jun 30 '16 at 02:35
  • 1
    It's worth moving the "My solution" section to an answer and then accepting it, because it means that the question will no longer appear in the "Unanswered" list. I think everyone who has looked at this question would agree. – bright-star Jan 29 '17 at 07:12
  • http://math.stackexchange.com/questions/2223400/how-to-evaluate-this-limit-riemann-integral look here – Sid Apr 09 '17 at 06:26

1 Answers1

3

This is a community wiki answer to remove this question from the unanswered list (once someone upvotes it): The OP's solution is correct, and very nicely written.

David E Speyer
  • 61,724
  • So we up vote your answer to do this? – zhw. Aug 29 '17 at 19:10
  • Right. Questions move off the unanswered list once they get an answer which either has a positive vote total or is accepted by the OP. (source https://meta.stackoverflow.com/questions/292013/the-unanswered-tab-has-multiple-answered-questions ). I always make answers community wiki when I do this so that I won't get the rep from your upvote. – David E Speyer Aug 30 '17 at 01:28
  • Done, thanks.$,$ – zhw. Aug 30 '17 at 02:52