Question

limit x (1- cos(1/ x)) approaches to minus infinity

Answer 1

\[\lim_{x\to-\infty}x\left(1-\cos\frac{1}{x}\right)=\lim_{x\to-\infty}\frac{1-\cos\dfrac{1}{x}}{\dfrac{1}{x}}=\frac{0}{0}\]
Apply L'Hopital's rule.
\[\lim_{x\to-\infty}\frac{-\left(-\dfrac{1}{x^2}\right)\left(-\sin\dfrac{1}{x}\right)}{-\dfrac{1}{x^2}}=\lim_{x\to-\infty}\sin\dfrac{1}{x}=\cdots\]