Show that the Function is not Continuous Show that the Function $$g(x):=\begin{cases} \left| x \right| \sin{( \cot{x} )} & \text{for }x\notin \left\{ 0,\frac{1}{42} \right\},  \\ 0 & \text{for }x=0, \\ 10^{42} & \text{for }x=\frac{1}{42} \end{cases}$$ 
is not continuous, but in the Point $\xi=0$ is continuous. You can use that $\sin,\cot$ are continuous functions without the need to prove it.

I have no idea how to show this. I would appreciate some help.
 A: To show continuity at $x=0$, you're being asked to show
$$
\lim_{x\to0} |x|\sin(\cot x) = \text{the given value} = 0.
$$
Since $\sin(\text{anything})$ is between $1$ and $-1$ (inclusive), you've got $|x|\sin(\text{something})$ between $|x|$ and $-|x|$, and those both approach $0$ as $x\to 0$, so that's all you need to do for that part.
For the other part, you need to show that
$$
\lim_{x\to1/42} |x|\sin(\cot x) \ne \text{the given value} = 10^{42}.
$$
Here it would suffice to show that


*

*$|1/42|\sin(\cot(1/42))$ is negative and

*since the function is continuous, it must remain negative in some neighborhood of $x=1/42$ and

*$10^{42}$ is positive.

A: Hint: What you have to prove is
$$\lim_{x\to\frac1{42}}|x|\sin(\cot x) \ne 10^{42} \\
\lim_{x\to 0}|x|\sin(\cot x) = 0\,.
$$
A: Hint: Try to prove the easy
Lemma: If for two functions $\,f(x)\,,\,g(x)\,$ we have that there exist $\,\epsilon\,,\, M\in\Bbb R_+\,$ s.t.:
$$\lim_{x\to x_0}f(x)=0\;\;\wedge\;\;|g(x)|\le M\,\,\,,\;\;\forall\,x\in (x_0-\epsilon,x_0+\epsilon)\;,\;\;\text{then also}\;\;\lim_{x\to x_0}f(x)g(x)=0$$
