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$$\lim _{x\to 0}\:\frac{\sin\left(\frac{1}{x^2}\right)}{x^2}$$

I my intuition is telling me this limit does not exist as $\sin$ will be oscillating but will stay bounded and then will blow up as $x \to 0$.

I can't seem to show it though. I tried to create two sequences that go to different limits but the $x^2$ in the numerator made this tricky so I'm not sure if that is intended?

Any help?

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One may consider, $\dfrac1{x_n^2}=\pi n$, with $n=1,2,3,\ldots$, giving

$$ \lim _{x_n\to 0}\:\frac{\sin\left(\frac{1}{x_n^2}\right)}{x_n^2}=\lim _{n\to +\infty}\pi n \times \sin(\pi n)=0 $$

and one may consider, $\dfrac1{y_n^2}=(4n+1)\dfrac{\pi}2 $, with $n=0,1,2,\ldots$, giving

$$ \lim _{y_n\to 0}\:\frac{\sin\left(\frac{1}{y_n^2}\right)}{y_n^2}=\lim _{n\to +\infty} (4n+1)\dfrac{\pi}2 \times \sin\left((4n+1)\dfrac{\pi}2\right)=\lim _{n\to +\infty}(4n+1)\dfrac{\pi}2 \times 1=\infty. $$

Thus the initial limit does not exist.

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  • $\begingroup$ Because this limit does not exist is it enough to prove $\lim _{x\to 0}\:\frac{x^6-\left(sin\left(\frac{1}{x^2}\right)\right)}{x^2}$? Also can't exist how can you state this in a good way mathematically speaking? $\endgroup$ – Robert May 25 '16 at 9:32
  • $\begingroup$ By just writing, $\frac{x^6-\sin\left(\frac{1}{x^2}\right)}{x^2}=x^4-\frac{\sin\left(\frac{1}{x^2‌​}\right)}{x^2}$ one of the two limits does exist, the other does not exist, thus the difference of two does not exist. Thanks. $\endgroup$ – Olivier Oloa May 25 '16 at 9:47
  • $\begingroup$ Olivier Oloa: Not true. Consider $f(x) = x^2 - \cos x $. $\lim_{x \to \infty} x^2 - \cos x = \infty $ however $\lim_{x \to \infty} \cos x $ does not exist... $\endgroup$ – user203867 May 25 '16 at 9:55
  • $\begingroup$ @Learner I don't see yo example as a counter example, since none of your two limits does exist... (maybe you have confused $x \to 0$ with $x \to \infty$) $\endgroup$ – Olivier Oloa May 25 '16 at 10:02
  • $\begingroup$ $\lim_{x \to \infty} x^2 = \infty $ $\endgroup$ – user203867 May 25 '16 at 10:04
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Let's write $t=\frac{1}{x^2}$, such that when $x\rightarrow 0$, then $t \rightarrow \infty$. Let's notice that $\lim_{x\rightarrow 0} \frac{\sin(\frac{1}{x^2})}{x^2}=\lim_{t\rightarrow \infty}t\sin(t)$, which doesn't exist and therefore the original limit doesn't exist.

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  • $\begingroup$ How would one show that the replaced limit does not exist ? $\endgroup$ – Robert May 25 '16 at 9:27
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    $\begingroup$ Let's take the following subseries: $a_n=2\pi n+\frac{\pi}{2}$. Now $\lim_{n\rightarrow \infty } (2\pi n+\frac{\pi}{2})sin(2\pi n+\frac{\pi}{2})=\infty$, but if we take $b_n=2\pi n$, then $\lim_{n\rightarrow \infty} (2\pi n)sin(2\pi n)=0$. $\endgroup$ – Noy Soffer May 25 '16 at 9:30
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Make the substitution $n = \frac{1}{X^2}$. Then it is clear that $n \to \infty \iff x \to 0 $. HEnce, we have

$$ \lim_{n \to \infty} n \sin n $$

Now, consider subsequences multiples of $pi$ and $pi/2$ and see that they converge to different limits...

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  • $\begingroup$ Got it thanks by the way because this limit does not exist is it enough to prove $\lim _{x\to 0}\:\frac{x^6-\left(sin\left(\frac{1}{x^2}\right)\right)}{x^2}$? Also can't exist $\endgroup$ – Robert May 25 '16 at 9:31

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