# Function defined as a limit

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 $[0,\pi/2]$.

• Nothing to add or rectify. +1.
– Did
Commented Nov 25, 2015 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! ^_^ Commented Nov 25, 2015 at 10:34
• 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. Commented Nov 25, 2015 at 17:02
• Well done. Just comment here so I can come back later :). Commented Jun 30, 2016 at 2:35
• 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. Commented Jan 29, 2017 at 7:12

• Done, thanks.$\,$