# Evaluate $\lim_{x \to 0} \frac{x^2}{x+\sin (\frac 1 x)}$

How can I find the following limit without the use of any series or expansion? $$\lim_{x \to 0} \frac{x^2}{x+\sin (\frac 1 x)}$$ Thanks for help.

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What are your thoughts? How do the various parts of the formula behave when $x$ becomes very small? –  Harald Hanche-Olsen May 2 '13 at 17:02
to zero but the way ? –  hmedan.mnsh May 2 '13 at 17:10
Note that this function isn't even defined for all $x$ near zero, since $x+\sin\frac{1}{x}=0$ is a clear possibility. –  Thomas Andrews May 2 '13 at 17:12
The limit does not exist. –  André Nicolas May 2 '13 at 17:12
for ex: xsin(1\x) when x >0 =0 so the limit are exist and we can use -x<xsin(1\x)<x .....because of lim x=lim -x=0 x>0 this lead to lim xsin(1\x) x>0 =0 is there any way like this to find the solution to my question –  hmedan.mnsh May 2 '13 at 17:18

As bob.sacramento suggests, let $y=1/x$ and then \begin{align} \lim_{x\to0}\frac{x^2}{x+\sin(1/x)} &=\lim_{y\to\infty}\frac1{y+y^2\sin(y)} \end{align} There are solutions near every non-zero integer multiple of $\pi$ for $y+y^2\sin(y)=0$. This can be easily verified since $y+y^2\sin(y)\gt0$ when $y=(2k+1/2)\pi$ and $y+y^2\sin(y)\lt0$ when $y=(2k-1/2)\pi$ for $k\ne0$.

For $y=k\pi$, $y+y^2\sin(y)=y$.

Thus, as $y\to\infty$, $\dfrac1{y+y^2\sin(y)}$ gets arbitrarily big and small. Thus, the limit does not exist.

Graphical Support

Here is a plot of $y+y^2\sin(y)$. It is easy to see why its reciprocal has no limit as $y\to\pm\infty$.

$\hspace{3cm}$

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Let $y=1/x$ and take the limit as $y\rightarrow \infty$.

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• Note that $x-1 \leq x+ \sin\frac{1}{x} \leq x+1$ and this says that $x^{2}$ increases much much faster than $x+\sin\frac{1}{x}$. So where should the limit go?
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Very good hint. +1 –  DonAntonio May 2 '13 at 17:07
for some strange reason, W|A gives the weird result +1 –  Santosh Linkha May 2 '13 at 17:08
@DonAntonio Well, Thank you! –  Seshadri May 2 '13 at 17:08
Actually, I'm not sure this hint leads in the right direction. There are values of $x$ arbitrarily close to zero where $x+\sin 1/x=0$. So I think there might not be a limit (even if we exclude those points.) –  Thomas Andrews May 2 '13 at 17:14
Just use Alpha, or some other tool, to find a solution of $\sin(1/x)=-x$ to $20$ decimal places. It is geometrically clear there are such. How perturb the $x$ a very tiny bit. –  André Nicolas May 2 '13 at 17:36

In the comments, several folks indicate that WolframAlpha produces strange results and/or should not be trusted too much. Of course, no software should be trusted too much but, in this particular case, I think that Alpha's response is perfectly reasonable. Here's what I see when I enter limit x^2/(x+sin(1/x)) as x->0:

Now as Don pointed out, Alpha reports that standard computation time was exceeded. Nonetheless, the pod of primary interest, namely the "Limit" pod, did complete and simply says "no result found...". This seems perfectly reasonable, although, "does not exist" or "undefined" could be even better.

As several folks also pointed out, an examination of a graph makes sense as well:

It now becomes crystal clear what's going on. For every $\varepsilon>0$ there are $x$ values with $0<|x|<\varepsilon$ and $x+\sin(1/x) = 0$ so that the function is simply not even defined on any entire punctured neighborhood of the origin.

Also, while it is certainly sufficient to show that the function is arbitrarily large and small in any neighborhood of the origin, it's not necessary to show this; really the problem is much simpler than that. Again, in order for the limit definition to be satisfied a function must be everywhere defined in some punctured neighborhood of the origin, i.e. some set of the form

$$\{x: 0<|x|<\varepsilon\}$$

and this function isn't - done. As a similar example, you might consider $f(x)=x\sin(1/x)/\sin(1/x)$. This function is, in fact, bounded and the same as $y=x$ except at an infinite sequence of points that converges to the origin where it is undefined. A graphical representation that somewhat captures this is the following:

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According to WA and to this site , the limit seems to exist and be equal to zero, yet in any neighbourhood of zero there is one point (well, in fact infinitely many) where the function's denominator vanishes and, thus, the limit cannot exist in fact. We can try to show this as follows: define

$$f(x)=x+\sin\frac1x$$

We now choose

$$(1)\;\;w_n:=\frac2{(4n+1)\pi}\;,\;n\in\Bbb Z\implies f(w_n)=\frac2{(4n+1)\pi}+1>0$$

$$(2)\;\;z_n:=\frac2{(4n+3)\pi}\;,\;n\in\Bbb Z\implies f(z_n)=\frac2{(4n+3)\pi}-1<0$$

Since clearly $\;w_n\,,\,z_n\xrightarrow[n\to\infty]{}0\;$ and since $\;f\;$ is continuous in the positive reals, we can see that in any neighborhood of zero $\,f\,$ changes sign infinite times, and this means our original function isn't defined in infinite points (using the MVT for continuous functions) within any "small" right neighbourhood of zero. The same can be shown on the negative side of zero.

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What limit "seems to exist and be equal to zero"? The original?? –  Mark McClure May 3 '13 at 0:02
Yes @MarkMcClure. In the links you can check it. It is pretty misleading. –  DonAntonio May 3 '13 at 0:04
Again - what is misleading. I'm honestly not sure what you're referring to. –  Mark McClure May 3 '13 at 0:13
The graph of the function $\,\frac{x^2}{x+\sin 1/x}\,$ is pretty misldeading, as it looks as if the limit exists, whereas, as pointed out, it doesn't... –  DonAntonio May 3 '13 at 0:25
Oh the graph. Hmm... It certainly doesn't look that way to me. Of course, I plotted just one graph from -0.1 to 0.1. In the course of numerical experimentation, one would naturally check the interval from -0.01 to 0.01 and see that the sequence of asymptotes continues. It's just a short step from there to examine the graph of both sides of $sin(1/x)=-x$ from which its geometrically clear that there are infinitely many solutions. –  Mark McClure May 3 '13 at 0:31
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note that this function is odd symmetric with respect to $x$. So if we can show that $\lim_{x\to 0^+}\frac{x^2}{x+\sin{1/x}}=0$, then $\lim_{x\to 0^-}\frac{x^2}{x+\sin{1/x}}=0$ as well. We get down to prove this.

Using squeeze theorem, we arrive at: $\lim_{x\to 0^+}\frac{x^2}{x-1}<\lim_{x\to 0^+}\frac{x^2}{x+\sin1/x}<\lim_{x\to 0^+}\frac{x^2}{x-0}$ and, $\lim_{x\to 0^+}\frac{x^2}{x+1}<\lim_{x\to 0^+}\frac{x^2}{x+\sin1/x}<\lim_{x\to 0^+}\frac{x^2}{x+0}$ and $\lim_{x\to 0^+}\frac{x^2}{x+1}=0$ ,also $\lim_{x\to 0^+}\frac{x^2}{x-1}=0$ ,and also $\lim_{x\to 0^+}\frac{x^2}{x+0}=0$, so

$\lim_{x\to 0^+}\frac{x^2}{x+\sin{1/x}}=0$, and using the symmetric property mentioned above, we get

$\lim_{x\to 0}\frac{x^2}{x+\sin{1/x}}=0$

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