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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?

Aronurr64
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    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 $\sin\frac 1x = 0$. – VIVID Apr 20 '21 at 19:09
  • Related – Theo Bendit Apr 20 '21 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. – GEdgar Apr 20 '21 at 21:46

1 Answers1

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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.

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    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 $x=0$. – Philipp Wacker Apr 20 '21 at 19:13
  • @MercuryBench: your example has little interest and is irrelevant to the question, sorry. When computing a limit, you never consider the function at the target point, that's well understood. The OP's question is quite different. –  Apr 20 '21 at 19:15
  • 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? – Aronurr64 Apr 20 '21 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. – Lubin Apr 20 '21 at 23:38
  • Could you clarify what you mean with "accumulation point" here? I looked it up but am having a hard time trying to understand what it means here. – Aronurr64 Apr 21 '21 at 00:29
  • @Aronurr64 $x$ is an accumulation point of a set $D$ if $x$ lies in the closure of $D \setminus {x}$. In other words, if there exists a sequence $(d_n)$ in $D \setminus {x}$ converging to $x$. Points are not accumulation points of $D$ when they lie a positive distance away from $D$, or if they lie in $D$ but are isolated. The idea is that $D$ clusters around $x$, so that limits may sensibly be defined. – Theo Bendit Apr 21 '21 at 06:10
  • @TheoBendit Thank you. How would you prove x to be an accumulation point? – Aronurr64 Apr 22 '21 at 15:11
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    @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 $x=0$. –  Apr 22 '21 at 15:14