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What is the value of the limit \(\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{sin^2x-1}{cos x}\)?(a) 0(b) 4(c) 1(d) Limit does not existI 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 |
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Answer» RIGHT choice is (a) 0 To elaborate: \(\lim\limits_{x \RIGHTARROW \frac{\PI}{2}}\frac{sin^2x-1}{cos x}\) = \(\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{-cos^2 x}{cos x}\) =\(\lim\limits_{x \rightarrow \frac{\pi}{2}}\) -COSX = 0 |
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