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My progress:

Using the sequential criterion for limits, I constructed two sequences $(x_n), (y_n)$ with $\lim(x_n)=\lim(y_n)=0$, such that $\lim(f(x_n))\neq \lim(f(y_n))$, where $f(x)=\sin\frac 1 x$.

So, $\lim_{x \rightarrow 0} \sin (\frac{1}{x})$ does not exist.

I also showed separately in the same way that $\lim_{x \rightarrow 0} \mathrm{sgn} (x)$ does not exist.

I know that $$\lim_{x \rightarrow 0} f(x)=M ,\, \lim_{x \rightarrow 0} g(x)=N \Rightarrow \lim_{x \rightarrow 0} (fg)(x)=MN$$

Here, I have two functions $f,g$ which do not have limits at $x=0$. Does it follow from here that $\lim_{x \rightarrow 0} (fg)(x)$ also doesn't exist?

Is it any other way to approach the problem?

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  • $\begingroup$ Look for the limits of $\frac{1}{x}$ from negative and positive sides. I think that argument can be easily used too and is very easy one. $\endgroup$ – Mann May 5 '15 at 12:46
  • $\begingroup$ Should I do that using $\epsilon-\delta$ definition? $\endgroup$ – Diya May 5 '15 at 12:46
  • $\begingroup$ Depends upon how much formal and rigorous you want it to be. $\endgroup$ – Mann May 5 '15 at 12:47
  • $\begingroup$ I need it to be formal but also easy, I'm a beginner at analysis.... :) Thanks! :) $\endgroup$ – Diya May 5 '15 at 12:48
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    $\begingroup$ @GFauxPas ah, ok, so it is not true! thank you :) $\endgroup$ – Diya May 5 '15 at 13:03
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Hint do you know for which $\theta$s you have $\sin \theta = 1$? how about $\sin \theta = -1$? Can you choose a sequence of $(x_n)$ and $(y_n)$ converging to 0 such that $\sin (1/x_n) = 1$ always and $\sin (1/y_n) = -1$ always?

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    $\begingroup$ sin$\theta=1$ for $\dfrac{(4n+1)\pi}{2}$ and sin$\theta=-1$ for $\dfrac{(4n-1)\pi}{2}$. So I will choose $(x_n)=\dfrac{1}{\frac{(4n+1)\pi}{2}}$ and $(y_n)=\dfrac{1}{\frac{(4n-1)\pi}{2}}$, right? Thanks a lot! :) $\endgroup$ – Diya May 5 '15 at 13:08
  • $\begingroup$ @Diya: That looks fine. $\endgroup$ – Willie Wong May 5 '15 at 13:12
  • $\begingroup$ Ok, so that proves that $lim_{x \rightarrow 0} sin(\frac1x)$ does not exist. How do I incorporate the $sgn$ part? $\endgroup$ – Diya May 5 '15 at 13:14
  • $\begingroup$ @Diya: the signum function $\mathrm{sgn}$ returns $+1$ if its argument is positive, and $-1$ if the argument is negative. In particular: $\mathrm{sgn}(+1) = +1$ and $\mathrm{sgn}(-1) = -1$. So for the sequences $(x_n)$ and $(y_n)$ that you constructed above, you have $\mathrm{sgn} \sin(x_n) = \sin(x_n) = +1$ and similarly for $y_n$. $\endgroup$ – Willie Wong May 5 '15 at 13:27
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    $\begingroup$ In fact, I am rather surprised that you were asked to show that this particular function has no limit at zero. If I were your instructor I would have assigned the two part question where the first part is to prove that for the function $f(x) = x \sin(1/x)$, the limit $\lim_{x\to 0^+} f(x)$ exists. And the second part to prove that $\lim_{x\to 0^+} \mathrm{sgn}\circ f(x)$ does not exist. This way you can see the difference that the signum function makes. $\endgroup$ – Willie Wong May 5 '15 at 13:30

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