calculus - By Cauchy's criterion of limit, show that $lim_{xto 0}(sin{frac{1}{x}}+xcos{frac{1}{x}})$ does not exist.

By Cauchy's criterion of limit (not sequencial criterion), show that

$$\lim_{x\to 0}(\sin{\frac{1}{x}}+x\cos{\frac{1}{x}})$$ does not exist.

Cauchy's criterion of limit

$\lim_{x\to c}f(x)=l$ iff for every $\epsilon>0$, there exists $\delta$ such that

$$|f(x_2)-f(x_1)|<\epsilon$$ for $0<|x_1-c|<\delta$ and $0<|x_2-c|<\delta$.

Please suggest $x_2, x_1$ and help me to solve the problem.

Answer

One part of Cauchy's Criterion says that

RESULT: If $\exists \epsilon>0$ such that $\forall \delta >0$, we can find $x_1,x_2$ satisfying $0<|x_1-a|<\delta$ and $0<|x_2-a|<\delta$ but $|f(x_2)-f(x_1)|\geq \epsilon$ then $\lim_{x\to a}f(x)$ does not exist.

Let us write

$$f(x)=\sin\frac{1}{x}+x\cos\frac{1}{x}.$$

Take $\epsilon\leq 2$. Let $\delta>0$. By Archimedean property,we can find $n\in\Bbb N$ such that $\frac{1}{n}<\delta$. Take

$$x_1=\frac{1}{\frac{3\pi}{2}+2\pi n}\qquad\text{and}\qquad x_2=\frac{1}{\frac{\pi}{2}+2\pi n}.$$ Notice that $0 $$|f(x_2)-f(x_1)|=|1-(-1)|=2\geq\epsilon.$$
Apply the result and we are done.

NOTE: That was the idea behind the hint given by @5xum. The only difference is that our delta and epsilon were interchanged. Since the OP want for clarification, I rather post an answer than to put all of these in the comments.