Question

Find \(lt_{x\rightarrow -5}(\frac{tan^{-1}(x^2+6x+5)}{x^2+15x+50})\)

(a) 0

(b) 1

(c) ^-4⁄5

(d) -1

The question was asked during an online interview.

This interesting question is from Indeterminate Forms in portion Differential Calculus of Engineering Mathematics

Answer 1

The correct choice is (c) ^-4⁄5

Easiest explanation: \(=lt_{x\rightarrow -5}\frac{tan^{-1}((x+5)(x+1))}{(x+5)(x+10)}\)

Now expand into Taylor Series for tan^-1(x)

\(=lt_{x\rightarrow -5}\frac{1}{(x+5)(x+10)}\times((x+5)(x+1)-\frac{(x+5)^3(x+2)^3}{3}+\frac{(x+5)^5(x+2)^5}{5}-..\infty)\)

\(=lt_{x\rightarrow -5}(\frac{(x+1)}{(x+10)}-\frac{(x+5)^2(x+2)^3}{3(x+10)}+\frac{(x+5)^4(x+2)^5}{5(x+10)}-…\infty)\)

\(=lt_{x\rightarrow -5}\frac{(-5)+1}{(-5)+10}\)

=\(-\frac{4}{5}\)