
pseudosine function?
My professor has this exercise in his notes. I may be able to prove it but I don't know what he pseudosine function is.
He states:
Let $\displaystyle \sigma$ be the pseudosine function. Let $\displaystyle f(x)=x\sigma (1/x)$ for $\displaystyle x\not= 0$. Prove $\displaystyle f$ has a removable discontinuity at $\displaystyle 0$.

The pseudosine function might be $\displaystyle \sigma(x) = \sin(\frac{1}{x})$, for $\displaystyle x\neq 0$.
In this case, we have $\displaystyle f(x) = x\sin(x)$ for $\displaystyle x\neq 0$, and it is not hard to check that this function has a removable discontinuity at $\displaystyle x=0$.
Do you know how to prove this?
Also, you better ask your professor, if this is the right definition of the pseudosine function (I recall having seen that name somewhere, definitely involving the function $\displaystyle \sin(\frac{1}{x})$, but it might have been $\displaystyle x\sin(\frac{1}{x})$ or something similar).

A more general concept of "pseudosine" is any function that is (quas) periodic and bounded. Of course, for this proof, you only need "bounded". If $\displaystyle a\le \sigma(x)\le b$ then $\displaystyle ax\le x\sigma(x)\le bx$. Then, as x goes to 0, $\displaystyle 0\le \lim_{x\to 0} x \sigma(x)\le 0$ so the limit exists and is 0. You only need "exists" to prove there is a removable discontinuity at x= 0.