# Using $\epsilon$-$\delta$ definition of limit to prove a limit doesn't exist.

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.

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

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.