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What is the value of the limit \(\lim\limits_{x \rightarrow 0}\frac{sin^2⁡x}{x^2}\)?(a) 2(b) 1(c) Limit does not exist(d) 4The question was asked in an interview for job.My question is based upon Limits of Trigonometric Functions in portion Limits and Derivatives of Mathematics – Class 11

Answer»

Correct CHOICE is (b) 1

Best explanation: \(\lim\limits_{X \rightarrow 0}\frac{sin^2⁡x}{x^2}\) =

= (\(\lim\limits_{x \rightarrow 0}\frac{sin ⁡x}{x}\) x \(\lim\limits_{x \rightarrow 0}\frac{sinx}{x}\))

We apply L’Hospital’s rule and DIFFERENTIATE numerator and denominator.

= (\(\lim\limits_{x \rightarrow 0}\frac{cos x}{1}\) x \(\lim\limits_{x \rightarrow 0}\frac{cos x}{1}\))

= 1



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