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Can someone explain, in simple terms, why the following limit doesn't exist?

$$\lim \limits_{x\to0}\sin\left(\left|\frac{1}{x}\right|\right)$$

The function is even, so the left hand limit must equal the right hand limit. Why does this limit not exist?

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5  
Graph it, or have software graph it for you. It will go nuts near $0$, wiggling desperately back and forth. –  André Nicolas Jul 13 '13 at 20:29
    
@ZettaSuro I think a way remove the wrong intuition that the limit exists because of the evenness of the function is to consider $\int \limits_{-\infty}^{+\infty}\sin (x)dx$. Two 'deconvergences' which are, in a sense, in accordance, do not turn into a convergence. –  Git Gud Jul 13 '13 at 20:36
1  
If the left limit exists, the right limit exists and is equal. –  Thomas Andrews Jul 13 '13 at 20:36

2 Answers 2

up vote 5 down vote accepted

The function is indeed even, as we will see, this does not prove the existence of the limit.

Assume that the limit exists.

$$ \lim_{x\to 0^+} \sin\left(\frac1x \right) $$

Which is the same as your limit because the left-hand limit and right-hand limit will be equal (if the limit exists). Then since $x > 0$, I removed the absolute value sign.

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

Using $u = 1/x$, we see that the limit certainly does not exist.

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Consider the following two sequences which both converge to $0$: $$ x_n:=\frac{1}{2n\pi}\qquad\text{and}\qquad y_n:=\frac{1}{\frac{\pi}{2}+2n\pi}. $$ If $f(x):=\sin(\lvert\frac{1}{x}\rvert)$, then we have $f(x_n)=\sin(2n\pi)=0$ for all $n$, and $f(y_n)=\sin(\frac{\pi}{2}+2n\pi)=1$ for all $n$. So, you have (very) different values in any arbitrarily small neighborhood of 0.

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