I am stuck trying to understand how to prove that the limit of $\sin(\frac{1}{x})$ as $x$ approaches $0$ does not exist. a hint was given: a limit does not exist if there exists an $\varepsilon > 0$ such that for every $\eta>0$ there exists an $x$ with $0<|x-a|<\eta$ whenever $|f(x)-L|\geq \varepsilon$.
so I approached this by stating that $|x|<\eta$ then $\sin(\frac{1}{x})-L\geq \varepsilon$. then I set $\varepsilon \leq|f(x)-L|<\sin(\frac{\eta}{2x})-L$ then used inverse sine to try and get delta alone… but now I'm looking at it trying to figure out what i did and why i did it.
can someone please walk me through a proof that this limit does not exist and maybe try and explain it to me. I am lost.
Best Answer
Hint: What are the values of the function at $\frac{2}{\pi}, \frac{2}{3\pi}, \frac{2}{5\pi}, \frac{2}{7\pi},\dots$?