Exercise 4, page 92 from Guidorizzi' book Calculo (in Portuguese) he asks to show that $\lim _{x\rightarrow 0}\sin\frac{1}{x}$ doesn't exist. I know how to do that by using two differents sequences, but that exercise is in the Squeeze Theorem section. So, is it possible to show that $\lim _{x\rightarrow 0}\sin\frac{1}{x}$ doesn't exist by using the Squeeze Theorem?
[Math] show that $\lim _{x\rightarrow 0}\sin\frac{1}{x}$ doesn’t exist by using the Squeeze Theorem
calculus
Best Answer
The usual squeeze theorem only says that a limit does exist, and it is not phrased as an "if and only if", so it can't literally be used to test for nonconvergence.
However, we can do something "squeeze theorem like". Given a limit $\lim_{x \to c} f(x)$, we can come up with "universal" functions $g(x)$ and $h(x)$, which take values in $[-\infty, \infty]$, such that for all $x$, $g(x) \leq f(x) \leq h(x)$, and such that the limit exists if and only if $\lim_{x \to c} g(x)$ and $\lim_{x \to c}h(x)$ both exist and are equal.
To make the notation easier, suppose $c = 0$. Both $g$ and $h$ will be even functions, so we only need to define them for non-negative $x$. We let $g(0) = -\infty$ and for $x > 0$ we let $$g(x) = \inf \{ f(y) : y \in [-x,0)\cup(0,x]\}$$
Similarly, we let $h(0) = \infty$ and for $x > 0$ we let $$h(x) = \sup \{ f(y) : y \in [-x,0)\cup(0,x]\}$$
Now, as we move towards $0$, $h(x)$ will decrease and $g(x)$ will increase, and so they will both have limits at $0$. If these limits are equal, then $g$ and $h$ can be used in the squeeze theorem to show that the limit of $f(x)$ exists. On the other hand, if the limits are not equal, then an argument directly using the definition of limits can be used to show the limit of $f(x)$ does not exist.
For $f(x) = \sin(1/x)$, we have that $g(x) = -1$ for all $x > 0$ and $h(x) = 1$ for all $x > 0$, so this method can be used to show that the $\lim_{x \to 0} \sin(1/x)$ does not exist.