Answer to Question #280103 in Real Analysis for abc

Question

Show that the function defined as 𝑓(𝑥) = { sin 1 𝑥 , 𝑥 ≠ 0 0, 𝑥 = 0 obeys the intermediate value theorem

Answer 1

The function $f(x)$ is continuous on $(-\infin, 0)\cup(0, \infin).$

So if $a$ and $b$ have the same sign then such $c$  exists by the Intermediate Value Theorem (IVT).

If $a\leq0\leq b.$  Assume $b>0.$

Let $n$ be a natural number such that $2/b <(2n+1)\pi$ so that the interval $J=[\dfrac{2}{(2n+3)\pi},\dfrac{2}{(2n+1)\pi}]$ is contained in $[0,b].$

The function $f(x) =\sin (1/x)$ is continuous on $J$ and takes on the values $1$ and $−1$ at the endpoints of $J.$

Since $−1 ≤ f(x) ≤ 1,$ the Intermediate Value Theorem applied to $f$ on $J$ implies that given any $y$ such that $−1 < y < 1,$ there is $c$ in $J$ such that $f(c) = y.$

The case $a<0$ is similar.