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What is the value of \(\lim\limits_{y \rightarrow 0}\frac{sin⁡3y}{3y}\)?(a) 0(b) 1(c) 3(d) \(\frac{1}{3}\)This question was posed to me during an interview.My question is taken from Limits of Trigonometric Functions in division Limits and Derivatives of Mathematics – Class 11

Answer»

Right answer is (B) 1

To elaborate: We KNOW that \(\lim\limits_{x \rightarrow 0}\frac{sin⁡x}{x}\) = 1.

Here x tends to 3Y.

Also, since this is of the form \(\frac{0}{0}\), we USE L’Hospital’s rule and differentiate the numerator and denominator separately.

= \(\lim\limits_{y \rightarrow 0}\frac{3\, cos\, 3y}{3}\)

= 1



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