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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.

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up vote 3 down vote accepted

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.
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I don't understand why is $|x| sin(something)$ between $|x|$ and $-|x|$ – Devid Mar 24 '13 at 15:40
Because the sine of any (real) number is between $-1$ and $1$. – Michael Hardy Mar 24 '13 at 15:45

Hint: What you have to prove is $$\lim_{x\to\frac1{42}}|x|\sin(\cot x) \ne 10^{42} \\ \lim_{x\to 0}|x|\sin(\cot x) = 0\,. $$

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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.:

$$\lim_{x\to x_0}f(x)=0\;\;\wedge\;\;|g(x)|\le M\,\,\,,\;\;\forall\,x\in (x_0-\epsilon,x_0+\epsilon)\;,\;\;\text{then also}\;\;\lim_{x\to x_0}f(x)g(x)=0$$

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