Question

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 1

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