[Math] $\lim_{x \to 0}\frac{\sin(\frac 1x)}{\sin (\frac 1 x)}$ ? Does it exist

limits

Does
$$\lim_{x \to 0}\;\frac{\sin\left(\frac 1x\right)}{\sin \left(\frac 1 x\right)}$$ exist?

I believe the limit should be $1$. Because function being defined at the point is not a condition for limit to exist.

This question came in my test and the answer given is limit does not exist.

But if we see the graph , it is quite clear the function is exact 1
as $x \to 0$, so the limit should be 0.

Even wolfram alpha gives the limit to be 1.

But we are playing with infinity, so who knows? Maybe I am missing out on something?

So what exactly is the limit and why?

Edit:

Wolfram alpha's widget (the link to which I have posted above) says the limit is 1.

But here wolfram alpha says that the limit doesn't exist on the real line.

Best Answer

I quote Walter Rudin's Principles of Mathematical Analysis for the definition of the limit of a function:

Let $X$ and $Y$ be metric spaces; suppose $E\subset X$, $f$ maps $E$ into $Y$, and $p$ is a limit point of $E$. We write $\lim_{x\to p}f(x)=q$ if there is a point $q\in Y$ with the following property: For any $\epsilon>0$, there exists a $\delta>0$ such that $d_Y(f(x),q)<\epsilon$ for all points $x\in E$ such that $0<d_X(x,p)<\delta$.

The symbols $d_X, d_Y$ refer to the distances in $X$ and $Y$, respectively.

In our case, $X=Y=\mathbb R$ with the metric $d(x,y)=|x-y|$. The function $f(x)=\frac{\sin \frac1x}{\sin \frac1x}$ maps the set $$ E=\mathbb R\setminus (\{\tfrac1{k\pi}:k\in \mathbb Z\setminus \{0\}\}\cup \{0\}) $$ into $\mathbb R$, and $0$ is a limit point of this set. We would conclude $\lim_{x\to 0}f(x)=1$ if for all $\epsilon>0$, we could find a $\delta>0$ so whenever $x\in E$ and $0<|x|<\delta$, then $|f(x)-1|<\epsilon$. But any $\delta$ suffices, since $f(x)=1$ for all $x\in E$.

Therefore, we do conclude that $\lim_{x\to 0}f(x)=1$.

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