# If $x_n=0$ infinitely often, does $\lim_{n\to \infty }\frac{y_n}{x_n}$ makes sense?

I have a sequence $$(x_n)$$ s.t. $$x_n=\frac{1}{n}$$ if $$n$$ even and $$0$$ if $$n$$ odd. We set $$y_n=\sin(x_n)$$. I have to compute $$\lim_{n\to \infty }\frac{y_n}{x_n}.$$

It's an exam question... intuitively, I would say that the limit is $$1$$, but does $$\lim_{n\to \infty }\frac{y_n}{x_n}$$ really make sense ? For me it make sense as far as $$y_n\neq 0$$ for all but finitely many $$n$$... What do you think ? But since it's an exam question, I'm probably wrong...

## 2 Answers

No, it does not make sense, because in order to talk about the limit of a certain sequence, say $$\{a_n\}$$, you need to be able to make a valid statement about the sequence for all but finitely many indices $$n$$, i.e. "...there exists $$N$$ such that for every $$n>N$$ the element $$a_n$$ satisfies such and such", which is obviously impossible if the sequence is undefined for infinitely many $$n$$'s.

As written it has no sense. To give a sense to your question : set $$f(x)=\begin{cases}1&x=0\\ \frac{\sin(x)}{x}&x\neq 0\end{cases}.$$ Then, indeed $$\lim_{n\to \infty }f(x_n)=1.$$ But, as written in your question I can't give any sense to it. Maybe other people will ?