Friday 25 October 2019

calculus - Limit of $sin(1/x)$ - why there is no limit?



$$ \lim_{x\to 0+} \sin\left(\frac{1}{x}\right)$$

I know that there is no limit.



but, why there is no limit?
I tried $x=0.4$, $x=0.3$, $x=0.1$, it looks like the limit is $0$.



And how can I show that there is no limit? I tried to calculate it like all the other functions, and I got wrong result and I don't know why:



$$\lim_{x \to 0+} \sin\left(\frac{1}{x}\right) = \sin\left(\frac{1}{0^+}\right) = \sin\left(\frac{1}{\infty}\right) = \sin(0) = 0.$$


Answer



Why there is no limit?




The graphic can help you understand why and suggest you some approach for the proof:



enter image description here



Remark: You have to be careful with tables of values because they can be misleading:



\begin{array}{ c | c c c c }
x & \frac{1}{2\pi} & \frac{1}{3\pi} & \frac{1}{4\pi} &\frac{1}{5\pi} \\ \hline
\sin\left(\frac{1}{x}\right) & 0 & 0 & 0 & 0 \\

\end{array}



\begin{array}{ c | c c c c }
x & \frac{2}{5\pi} & \frac{2}{9\pi} & \frac{2}{13\pi} &\frac{2}{17\pi} \\ \hline
\sin\left(\frac{1}{x}\right) & 1 & 1 & 1 & 1 \\
\end{array}



(The tables above are a sketch of the proof - see Theorem 2.4 here.)


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...