Question

Find \(lt_{x\rightarrow -101}(\frac{ln(x^2+20x+(x+101)^2(x^2+3))-ln(x^2+20x)}{x^2+202x+10201})\).

(a) 0

(b) 1

(c) ∞

(d) \(=\frac{10204}{12221}\)

I had been asked this question during a job interview.

I need to ask this question from Indeterminate Forms topic in portion Differential Calculus of Engineering Mathematics

Answer 1

Correct option is (d) \(=\frac{10204}{12221}\)

Best explanation: \(=lt_{x\rightarrow -101}(\frac{ln(1+\frac{(x+101)^2(x^2+3)}{(x^2+20x)}}{(x+101)^2})\)

Now expanding into Taylor series we have

\(=lt_{x\rightarrow -101}(\frac{1}{(x+101)^2})\times(\frac{(x+101)^2(x^2+3)}{(x^2+20x)}-\frac{(x+101)^4(x^2+3)^2}{2(x^2+20x)^2}+…\infty)\)

\(=lt_{x\rightarrow -101}(\frac{(x^2+3)}{(x^2+20x)}-\frac{(x+101)^2(x^2+3)^2}{2(x^2+20x)^2}+….\infty)\)

All others exceptthe first term tend to zero. Thus, we have

\(=lt_{x\rightarrow -101}(\frac{(-101)^2+3}{(101)^2+20(101)})\)

\(=\frac{10204}{12221}\)