Question

Find \(lt_{x\rightarrow -2}\frac{sin(\frac{1+(\frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2})}{(x+2)}\)

(a) ∞

(b) 0

(c) 2

(d) -∞

This question was addressed to me at a job interview.

My doubt stems from Indeterminate Forms topic in division Differential Calculus of Engineering Mathematics

Answer 1

Correct answer is (c) 2

For explanation I would say: First evaluate

\(=lt_{x\rightarrow -2}\frac{ln(1+\frac{(x+2)^2(x^2+1)}{x^3+3})}{x+2}\)

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

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

We hence the form for the totle limit as

\(lt_{x\rightarrow a}\frac{sin(f(x))}{g(x)}\)=1

Where f(x)→0 : g(x)→0 as x→a

This is true for the above problem

Thus, we can deduce the limit as

 = 1

Hence, 2 is the right answer.