Vtyjkyuk

Here's my attempt:

Suppose there is some $L in mathbbR$ such that $lim_ xto 0 sin frac2pi x = L$. Then, if we let $varepsilon = 1$, there would exist $delta > 0$ such that $left| f(x) - Lright| < 1$ for all $0< |x| < delta $. Now, by the Archimedean property, there is $n in mathbbN$ such that $frac1n < delta $. Pick $x_1 = frac44n+1$ and $x_2 = frac44n+3$ and note that $x_1 , x_2 < delta $. Now for $x_1$, we have

$ left| f(x_1) - L right| = left| sin left( (4n+1) fracpi2 -L right) right| = left| 1-L right| < 1 $

And similarly

$ left| f(x_2) - L right| = left| sin left( (4n+3) fracpi2 -L right) right| = left| -1-L right| < 1 $

Thus, L is within one unit of both $-1$ and $1$. However, there cannot be any such number. A contradiction!

Is this proof okay?

The only mistake I can find is that $x_1 , x_2 < delta$ ought to be $0<x_1 , x_2 < delta$. Also, depending on who's grading, "there cannot be such a number" might need some proof or justification. Or it might not. Apart from that, well done!

As an alternative simply note that for $nto infty$

• $x_n=frac4n to 0 implies sin frac2pifrac4n=sin fracpi n2=pm 1$

therefore the given limit doesn’t exist.