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Do limit exist for the following functions:

  1. $\lim_{x\rightarrow 0} \cos(\frac{1}{x})$

I think it exists because the expression for Left Hand Limit & Right Hand Limit are same

i.e $\lim_{h\rightarrow 0} \cos(\frac{1}{h})$ for $x=0+h$ & $x=0-h$

  1. $\lim_{x\rightarrow 0} \sin(\frac{1}{x})$

I think the limit doesn't exist because the expression for Left Hand Limit & Right Hand Limit are different.

$LHL:-\lim_{h\rightarrow 0} \sin(\frac{1}{h})$ for $x=0-h$

and

$RHL:\lim_{h\rightarrow 0} \sin(\frac{1}{h})$ for $x=0+h$

Is my thought process correct?


If the above is correct, then can we say the following with similar arguments:

(i)Limit exists for $RHL:\lim_{x\rightarrow 0} \dfrac{1}{| x |}$

(ii)Limit doesn't exist for $RHL:\lim_{x\rightarrow 0} \dfrac{1}{x}$

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  • $\begingroup$ From the answers given below it seems the limit doesn't exist because we can't find it. Is this argument valid for $\dfrac{1}{| x |}$ & $\dfrac{1}{x}$ $\endgroup$ – kryptoknight Aug 11 '13 at 20:07
  • $\begingroup$ I'm not sure what you mean by "this argument", but neither the limit of $\frac1x$ nor of $\frac1{|x|}$ as $x\to 0$ exists. However, you might say that $\lim_{x\to0}\frac1{|x|}=\infty$. The limit still doesn't "exist", but this is a more precise way of describing the behavior at $x=0$ $\endgroup$ – Omnomnomnom Aug 12 '13 at 0:10
  • $\begingroup$ @Omnomnomnom:By 'this' i mean non existence of limit because it comes out to be $\infty$ which can't be defined. $\endgroup$ – kryptoknight Aug 12 '13 at 17:21
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Actually, for both functions $\cos\left(\frac 1x\right)$ and $\sin \left(\frac 1x\right)$, the limits as $x\to 0$ do not exist, and for the same reasons, irregardless of whether we consider the limit as $x \to 0^+$ or $x\to 0^{-1}$.

Look, for example, of the behavior of $\cos \left(\frac 1x\right)$ on the interval $(-0.1, 0.1)$.

enter image description here

For any function $f(x)$, for a limit $L$ to exist as $x \to a$, we have $$\lim_{x\to a}f(x)=L\in \mathbb R\iff \forall(x_n)\to a,\; f(x_n)\to L$$

For both $f(x) = \cos \frac 1x$ and $f(x) = \sin \frac 1x$, there is no such $L$ to which $f(x)$ as $x \to 0$ from the right OR from the left.

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  • $\begingroup$ Pictures help so much! +1 $\endgroup$ – Amzoti Aug 12 '13 at 0:06
  • $\begingroup$ I love this function because of puzzling plot. Have a good and deep slumber Amy. :) $\endgroup$ – mrs Dec 3 '13 at 5:26
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Recall that $$\lim_{x\to a}f(x)=\ell\in\overline{\mathbb R}\iff \forall(x_n)\to a,\; f(x_n)\to\ell$$ so take $$x_n=\frac{1}{n\pi}$$ to show that the limit doesn't exist.

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You are correct in saying that since $\cos\left(\frac1x\right)$ is an even function: if the limit as $x\to0$ from one direction exists, it must be equal to the limit from the other direction. However, neither $\lim_{x\to0^+}\cos\left(\frac1x\right)$ nor $\lim_{x\to0^-}\cos\left(\frac1x\right)$ exist. In fact, these limits don't exist for the same reason that $\lim_{x\to0^+}\sin\left(\frac1x\right)$ doesn't exist, and comparing the right-side and left-side limits doesn't give you the whole picture in any of these cases.


Here's another way of thinking about this question to hopefully put you on the right track: what is the limit $\lim_{x\to\infty}\sin(x)$? Does it exist? If so, what is it? If not, then why not? Why is this limit the same as your limit?

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