Prove using Cauchy criterion that the limit of the function $\cos(\frac{1}{x})+x^2$ doesn't exist. (when $x_0 \rightarrow 0$) so we need to prove that:
Exist $\epsilon \ge 0$ such that for all $\delta \gt 0$ exist $x,y$ such that $0 \lt |x-x_0| \lt \delta \wedge 0 \lt |y-y_0| \lt \delta$ And $|f(x)-f(y)| \ge \epsilon$
Well I tried using it but failed in the middle , I don't know how to proceed.
My try :
$|f(x)-f(y)|=|\cos(\frac {1}{x})+x^2 - \cos(\frac {1}{y})-y^2| \ge ?$
Thank you in advance.
Take $x_0,y_0 = 0$. Let $\epsilon = 2$, and $\delta>0$. By the Archimedian property of $\mathbb R$, there exist $n,m\in\mathbb N$ such that $$x = \frac{1}{2n\pi}<\delta,\qquad\text{and}\qquad y = \frac{1}{-2m\pi + \pi}<\delta.$$ Without loss of generality, we can assume further that $x<y$. Since $\cos(x^{-1}) = 1$, and $\cos(y^{-1}) = -1$, it follows that $$\begin{align*}\vert f(x) - f(y)\vert &= \vert\cos(x^{-1}) - \cos(y^{-1}) + x^2 - y^2\vert \\ &= \vert 2 + x^2 - y^2\vert \\&> 2 = \epsilon. \end{align*}$$