# Does $\lim\limits_{x\to0}x^2\csc\frac{1}{\sqrtx}$ exist?

The original problem is computing the limit $$L=\lim_{x\to1}\frac{(x-1)^2}{\sin\frac{1}{\sqrt{x-1}}}$$ for which I replaced $x-1$ with $x$.

Is there something wrong with invoking the limit $$\lim_{x\to0}\frac{\sin x}{x}=1$$ or rather, $$\lim_{x\to0}\frac{x}{\sin x}=1$$ in this case? In doing so, I have $$L\stackrel?=\lim_{x\to0}\frac{x^2}{\sin\frac{1}{\sqrtx}}\times\frac{\frac{1}{\sqrtx}}{\frac{1}{\sqrtx}}\stackrel?=\lim_{x\to0}\sqrt{x^7}=0$$ yet a plot (over $-10^{-2}$ and $+10^{-2}$ below) in Mathematica suggests the function oscillates between positive and negative infinity. • $$\lim_{x\to0}\frac{x}{\sin x}=1 \implies \lim_{x\to \infty}\frac{x^{-1/3}}{\sin (x^{-1/3})}=1$$ you used that $\lim_{x\to 0}\frac{x^{-1/3}}{\sin (x^{-1/3})}=1$ which is not correct
– WW1
Feb 10 '16 at 22:06
• The limit is $0$! Feb 10 '16 at 22:07
• Consider that $\sin (x^{-1/3})$ becomes $0$ frequently as $x\to0^+$...
– user228113
Feb 10 '16 at 22:28

We are trying to compute $$\lim_{x \to 0} \frac{x^2}{\sin\left(\frac{1}{x^{1/3}}\right)}.$$ The numerator goes to zero, but is nonzero for $x \neq 0$. The function $\frac{1}{x^{1/3}}$ goes to infinity as $x \to 0$, and so $\sin(\frac{1}{x^{1/3}}) = 0$ for infinitely many $x$ in any neighborhood of $0$. Near these zeroes, the original function $x^2 / \sin(x^{-1/3})$ is near $\pm \infty$.
Thus your suspicion from Mathematica is correct --- the function does have infinitely many $\pm \infty$ oscillations, and the limit does not exist.
When $n\in \mathbb N$ and $x=1/(\pi n)^3$ the denominator is $0$ in $$f(x)=x^2/\sin (1/x^{1/3}).$$ Even if we exclude such values, the limit does not exist.
For $n\in \mathbb N$ let $x_n=1/(\pi n)^3.$ Let $0<d_n<x_n/2$ where $d_n$ is small enough that $x_n/2<y_n=1/(\pi n+d_n)^3$ and $z_n=1/(\pi n-d_n)^3<2x_n$ and $0<\sin d_n<x_n^2/n.$
Then $0<y_n<x_n<z_n<2x_n,$ so $y_n$ and $z_n$ go to $0$ as $n\to \infty.$ And $f(y_n)<-n/4$ while $f(z_n)>n.$