Limits paradox in case of sine function What according to you will be the answer for:
$$\lim_{x \to 0} \frac{\sin(1/x)}{\sin(1/x)}$$
My answer was $1$. The book says not defined and the argument is that both numerator and denominator are in $[-1,1]$, so we cannot determine what it exactly is.
My argument is, whatever value it takes, shouldn't it be the same in both numerator and denominator? Moreover, another example may be:
$$\lim_{x \to 0} \frac{1/x}{1/x}$$
I say $1$ again, because it is the same. But, it is not defined if we individually consider numerator and denominator! Still we have cancelled it in this case. Why can't I do it in case of sine function?

I don't know if it should be a comment or an answer:
In most books, the definition of limit is: given $ \epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ for all $0<|x| < \delta$. Using this definition, there is no limit, as explained by many other answers.
However, some books use: given $ \epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ for all $x$ in the domain of $f$ satisfying $0<|x| < \delta$. With this second definition, then your function is constant on its domain of definition, and the limit exists.
I guess your book is using the first definition.  
