Question

Find \(lt_{x\rightarrow 0}\frac{sin((4x^3)tan^{-1}(x))}{x^4}\)

(a) 1

(b) 2

(c) 4

(d) 3

I have been asked this question during an interview.

My question is taken from Indeterminate Forms in division Differential Calculus of Engineering Mathematics

Answer 1

The correct answer is (c) 4

To elaborate: Expand into Mclaurin series

\(=lt_{x\rightarrow 0}(\frac{1}{x^4})\times(\frac{4x^3 tan^{-1}(x)}{1!}-\frac{16x^6 tan^{-1}(x)}{3!}+…\infty)\)

\(=lt_{x\rightarrow 0}(\frac{4 tan^{-1}(x)}{x1!}-\frac{16x^2 tan^{-1}(x)}{3!}+…\infty)\)

Neglecting higher order terms (which go to zero) we have

\(=lt_{x\rightarrow 0}(\frac{4 tan^{-1}(x)}{x!})=lt_{x\rightarrow 0}(\frac{4}{x})\times(\frac{x}{1}-\frac{x^3}{3}+..\infty)\)

\(=lt_{x\rightarrow 0}(\frac{4 tan^{-1}(x)}{x!})=lt_{x\rightarrow 0}(\frac{4}{1}-\frac{4x^2}{3}+…\infty)=4\)