Question

What is the value of the \(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{sin⁡2x}\)?

(a) \(\frac{-1}{2}\)

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

(c) \(\frac{1}{4}\)

(d) \(\frac{-1}{4}\)

I got this question during an internship interview.

The question is from Limits of Trigonometric Functions topic in chapter Limits and Derivatives of Mathematics – Class 11

Answer 1

Correct option is (b) \(\frac{1}{2}\)

To explain I would say: \(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{sin⁡2x}\) =\(\lim\limits_{x \rightarrow \frac{3\pi}{2}}\frac{cos⁡ x sin⁡ x}{2 cos x sin⁡ x}\)

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