# Proving $\lim\limits_{x\to0}{\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})}}=1$

I am relatively new to calculus and I'm trying to understand it rigorously. For this question, assume that I only consider the functions to have purely real domains and ranges.

I have seen that $$\lim\limits_{x\to0}{\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})}}=1$$ is considered true. This is what my calculator and an online limit solver gave, so unless they are wrong, there must be a flaw in my reasoning. I do not understand it or know how to prove it using epsilon and delta. Here's what I've thought so far:

For any function f, $$\frac{f(x)} {f(x)}=1$$ if f is nonzero and defined at x. If these conditions are met when $$|x-c|\in(0, \delta)$$, the quotient is one, so any epsilon satisfies the condition. This means that to prove $$\lim\limits_{x\to c}{\frac{f(x)}{f(x)}}=1$$, you just need to show that f is nonzero and defined when $$|x-c|\in(0, \delta)$$. I don't believe this can be shown for $$f(x)=\sin(\frac{1}{x})$$, since it has infinitely many x-intercepts within any $$\delta$$.

Have I made a mistake anywhere or failed to consider something? How would this be normally demonstrated?

• You need to show for all $x$ such that $0 < x < \delta$, the following inequality holds:$$\left|\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})}-1\right| < \varepsilon$$ excluding $x$ where $\sin\frac 1x = 0$. Commented Apr 20, 2021 at 19:09
• Related Commented Apr 20, 2021 at 19:19
• Or: the zeros of $\sin(1/x)$ are removable discontinuities of $\sin(1/x)/\sin(1/x)$, so in some circumstances, we consider this modification. Maybe your calculator is using a different convention than your calculus text. Commented Apr 20, 2021 at 21:46

The "trick" is that you only consider evaluations of the function in its domain. The point where you want to compute the limit ($$x=0$$) needs to be an accumulation point, and it is the case here. As for all values in the domain the ratio is $$1$$, this is indeed the value of the limit.
• As a concrete example: We can consider $\lim_{x\to 0} \frac{\frac{1}{x}}{\frac{1}{x^2}}$, although both numerator and denominator are ill-defined at $x=0$. Commented Apr 20, 2021 at 19:13
• So do I have the wrong understanding of the epsilon-delta definition? Does the definition only require that the function falls within $\varepsilon$ of the point when the function is defined? Commented Apr 20, 2021 at 19:18
• When I was teaching, I emphasized that in a limit $\lim_{x\to a}f(x)$, you do not care what the value of $f(a)$ is, whether defined or not. Commented Apr 20, 2021 at 23:38
• @Aronurr64: the philosophy of a limit is that you can obtain values of $f$ as close as you want to the target point. Ponder this in the case of $\sin\frac1x$ around $x=0$.