Intuition for limit not existing Consider the function:
For $x \neq 1$, $f(x) = 2(x-1)\cos(\frac{1}{x-1}) + \sin(\frac{1}{x-1})$,
and for $x = 1$, $f(x) = 0$.
Why does the limit of $f(x)$ as $x \to 1$ not exist intuitively speaking? How can one tell?
 A: When $x$ is very close to $1$, $(x-1)\cos\left(\frac1{x-1}\right)$ is very close to $0$. But $\sin\left(\frac1{x-1}\right)$ keeps moving between $1$ and $-1$ as $x$ approaches $1$.
A: Consider the simpler example of the function
$$
\phi(x)=\sin\left(\frac{1}{x}\right) \, ,
$$
which displays roughly the same oscillatory behaviour as the function you are describing.

No matter how close $x$ is to $0$, $\phi(x)$ could be equal to anything between $-1$ and $1$. Hence, the limit does not exist.
Now consider the function
$$
\psi(x)=x\sin\left(\frac{1}{x}\right) \, .
$$

Although $\psi(x)$ is not defined when $x=0$, when $x$ is close to $0$, $\psi(x)$ is close to $0$. Think of a tiny number, say $0.001$. Provided that $x$ is sufficiently close to $0$, $-0.001<\psi(x)<0.001$. Indeed, for any positive number $\varepsilon$, it will be the case that
$$
-\varepsilon < \psi(x) < \varepsilon
$$
provided that $x$ is sufficiently close to, but unequal to, $0$.
A: Since $2(x-1)$ goes to $0$ as $x\to 1$ the first part goes to $0$ but $\sin(1/1-x)$ keeps oscillating between $-1$ and $1$ this is because the function $\sin$ is periodic and $1/1-x$ always hits the pace where sin becomes $1$ and $-1$ (and everything in between).
