Question

What is the value of \(\lim\limits_{x \rightarrow \infty}\frac{x sin⁡\frac{2}{x}}{2}\)?

(a) 1

(b) 2

(c) \(\frac{1}{2}\)

(d) Limit does not exist

The question was posed to me in an international level competition.

I want to ask this question from Limits of Trigonometric Functions in portion Limits and Derivatives of Mathematics – Class 11

Answer 1

Right choice is (a) 1

To explain: This is of the form \(\frac{0}{0}\), so we use L’Hospital’s rule.

= \(\lim\limits_{x \rightarrow \infty}\frac{\frac{-2}{x^2}cos⁡\frac{2}{x}}{\frac{-2}{x^2}}\)

= \(\lim\limits_{x \rightarrow \infty}\)cos\(\frac{2}{x}\)

= 1