Question

What is the value of the limit \(\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{sin^2⁡x-1}{cos ⁡x}\)?

(a) 0

(b) 4

(c) 1

(d) Limit does not exist

I have been asked this question by my college professor while I was bunking the class.

My enquiry is from Limits of Trigonometric Functions in chapter Limits and Derivatives of Mathematics – Class 11

Answer 1

Right choice is (a) 0

To elaborate: \(\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{sin^2⁡x-1}{cos ⁡x}\) = \(\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{-cos^2 x}{cos ⁡x}\)

=\(\lim\limits_{x \rightarrow \frac{\pi}{2}}\) -cosx

= 0