There are two definitions on the limit of a function:
Definition 1: Let $f(x)$ be a function defined on a neighbourhood of $x_0$, $\lim_{x\to x_0} f(x)=A $ if only if for every real $\varepsilon>0$, there exists a real $ \delta>0$ such that for all $\mathrm{x}, 0<|x-x_0|<\delta$ implies that $|f(x)-A|<\varepsilon$.
Definition 2:
Apart from open intervals, limits can be defined for functions on arbitrary subsets of $\mathbf{R}_{\text {, }}$ as follows (Bartle & Sherbert 2000): let $f$ be a real-valued function defined on a subset $S$ of the real line. Let $p$ be a limit point of $S$-that is, $p$ is the limit of some sequence of elements of $S$ distinct from $p$. The limit of $f_{t}$ as $x$ approaches $p$ from values in $S_{t}$ is $L$, if for every $\varepsilon>0$, there exists a $\delta>0$ such that $0<|x-p|<\delta$ and $x \in S$ implies that $|f(x)-L|<\varepsilon$.
Let $f(x)=\frac{\sin(x^2\sin \frac{1}{x})}{x^2\sin \frac{1}{x}}$, Obviously, $f(x)$ is not defined on $x=\frac{1}{n\pi}, n\in Z$ and $x=0$.
I am wondering whether $\displaystyle\lim_{x\to 0} \frac{\sin(x^2\sin \frac{1}{x})}{x^2\sin \frac{1}{x}}$ exists.
I think the above limit exists (equals to 1) according to definition 2 but not according to definition 1. Any help would be appreciated!
