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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...
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 ?
