Question

What is the value of the limit f(x) = \(\frac{sin^2⁡x+\sqrt 2 sin ⁡x}{x^2-4x}\) if x approaches 0?

(a) \(\frac{1}{\sqrt 2}\)

(b) \(\frac{-1}{\sqrt 2}\)

(c) \(\frac{-1}{2\sqrt 2}\)

(d) \(\frac{1}{2\sqrt 2}\)

I have been asked this question in an interview for internship.

This is a very interesting question from Limits of Trigonometric Functions topic in chapter Limits and Derivatives of Mathematics – Class 11

Answer 1

Correct answer is (c) \(\frac{-1}{2\sqrt 2}\)

The best I can explain: This is of the form  \(\frac{0}{0}\), therefore we use L’Hospital’s rule and differentiate the numerator and denominator.

= \(\lim\limits_{x \rightarrow 0}\frac{2sin⁡ \,x cos \,⁡x + cos \,⁡x \sqrt 2}{2x – 4}\)

= \(\frac{0+\sqrt 2}{-4}\)

= \(\frac{-1}{2\sqrt 2}\)