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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 existThe 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»

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



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