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How can I use the $\epsilon$-$\delta$ definition of limit to prove that the following limit doesn't exist? $$ \lim_{x\to1} \sin(\frac{1}{x-1}) $$ So far, I have tried to write out the definition of limit and what we get from that, but I am afraid that I haven't made much progress. My work so far:

Assuming the contrary, let's say the limit exists and is equal to L. From the definition of limit we get that $0<\lvert{x-1}\rvert<\delta$ and hence we have to find some $\delta$ which will imply $\lvert\sin(\frac{1}{x-1})-L\rvert <\epsilon$. I am not able to get beyond this, I was thinking about getting a lower bound on $\sin(\frac{1}{x-1})$ using the first inequality, but I doubt that's going to help in constructing a counter-example.

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    $\begingroup$ Hint: For any value $c$ in $[-1, 1]$, there are $x$-values arbitrarily close to $x = 1$ such that $\sin(1/(x - 1))$ equals $c$. $\endgroup$
    – LSpice
    Commented Mar 10, 2022 at 3:35
  • $\begingroup$ Can you expand a little bit on your hint ? I don't seem to understand how to use it in my proof. $\endgroup$
    – Italian
    Commented Mar 10, 2022 at 3:42
  • $\begingroup$ That's how you contradict $\lvert\sin(1/(x - 1)) - L\rvert < \epsilon$, after choosing $\epsilon$ small enough. $\endgroup$
    – LSpice
    Commented Mar 10, 2022 at 3:53
  • $\begingroup$ Can you please show me a complete proof if you have the time? I'm sorry but I can't seem to understand how to do it. $\endgroup$
    – Italian
    Commented Mar 10, 2022 at 4:03

1 Answer 1

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How can I use the $\epsilon$-$\delta$ definition of limit to prove that the following limit doesn't exist? $$ \lim_{x\to1} \sin(\frac{1}{x-1}) $$

Let $~\displaystyle f(x) = \sin\left(\frac{1}{x-1}\right).$

I guess that different people attack this problem in different ways. My approach is to establish that no matter how small a neighborhood of $\delta > 0$ is taken around $x = 1$, I will always be able to find distinct values $x_1, x_2$ that are both inside this neighborhood, so that (for example), for a fixed $r > 0$, you have that

$$|f(x_1) - f(x_2)| > 2r.$$

Assume that this has been done. Then, set $\epsilon = r$, and consider whether the function can converge to a limit $L$. The problem is that

$$|f(x_1) - L| + |f(x_2) - L| > 2r = 2\epsilon, \tag1$$

by the triangle inequality. Therefore, in (1) above, at least one of the two LHS terms must be greater than $\epsilon$. Further, by presumption, this will hold no matter how small $\delta$ is taken. This implies that it is impossible for any limit $L$ to exist such that the function converges to $L$.


Therefore, the problem has been reduced to establishing that regardless of how small $\delta > 0$ is taken, I can find $x_1, x_2$ as distinct values such that

  • $0 < |x_1 - 1| < \delta.$
  • $0 < |x_2 - 1| < \delta.$
  • $|f(x_1) - f(x_2)| > (1/2)$ (for example).

Assuming that the above is demonstrated, then (for example) I could take $\epsilon = (1/4)$, and then apply the analysis of the previous section.


For any (fixed) $~\delta > 0,~$ choose $~M \in \Bbb{Z^+},$ such that $\displaystyle ~M > \frac{1}{\delta}.$

Then, set

  • $~\displaystyle x_1 = \frac{1}{M\pi} + 1 \implies $ $\displaystyle |x_1 - 1| = \frac{1}{M\pi} < \frac{1}{M} < \delta.$
  • Similarly, set $~\displaystyle x_2 = \frac{1}{[M+(1/2)]\pi} + 1.$

So, now, $x_1, x_2$ are distinct elements in a neighborhood of $\delta$ around $x=1$.

Then

  • $\displaystyle \frac{1}{x_1 - 1} = M\pi \implies f(x_1) = 0.$
  • $\displaystyle \frac{1}{x_2 - 1} = [M + (1/2)]\pi \implies f(x_2) = \pm 1.$

Thus, $~|f(x_1) - f(x_2)| > (1/2),~$ as required.

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