# show that $\lim(\sin(\frac{1}{x}))$ as $x$ approaches zero does not exist.

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.

• I'm sorry, in the first paragraph it should say that there exists an x with 0<|x-a|<delta when |f(x)-L|>=epsilon Aug 25 '14 at 18:34
• Actually, it should say that for every $\epsilon>0$ there exists $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$. You should start by drawing a graph of the function and understanding why the limit does not exist. Aug 25 '14 at 18:35
• You can also use $\lim_{x\to 0^+}\sin(\frac{1}{x}) = \lim_{x\to +\infty}\sin(x)$ Aug 25 '14 at 18:53

Hint: What are the values of the function at $\frac{2}{\pi}, \frac{2}{3\pi}, \frac{2}{5\pi}, \frac{2}{7\pi},\dots$?
Proof by contradiction. Suppose $\lim_{x \to \infty} \sin\frac{1}{x}$ exists and it's equal $a$. Then exists such a $\delta$ for which $|a-x|<\frac{1}{4}$ when $x < \delta$. But there exists $k$ for which $\frac{1}{2k\pi}<\delta$ and $\frac{1}{2k\pi+\frac{\pi}{2}}<\delta$. Next:
$$\sin\frac{1}{\frac{1}{2k\pi}}=\sin 2k\pi=0$$
$$\sin\frac{1}{\frac{1}{2k\pi+\frac{\pi}{2}}}=\sin (2k\pi+\frac{\pi}{2})=1$$
Because $\frac{1}{2k\pi}<\delta$ and $\frac{1}{2k\pi+\frac{\pi}{2}}<\delta$ you have $|a-1|=|1-a|<\frac{1}{4}$ and $|a-0|=|a|<\frac{1}{4}$. But by triangle inequality:
$$1=|1|<|a|+|1-a|<\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$