Question

Find lt x → ∞\((\frac{ln(1+\frac{(x+3)^3(2x+9)}{(4x^3+3)})}{x^3+3x^2+9x+27})\)

(a) 0

(b) 1

(c) Undefined

(d) – ^1⁄35

This question was addressed to me in class test.

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

Answer 1

Correct option is (d) – ^1⁄35

The best I can explain: The form here is of ^0⁄0

Applying L hospitals rule would be really tough to differentiate. Hence we use the concept of Taylor Series

We know that ln(1+x)=\(x-\frac{x^2}{2}+\frac{x^3}{3}-…\infty\)

Thus, we have

\(=lt_{x\rightarrow -3}(\frac{1}{x^3+3x^2+9x+27}\times (\frac{(x+3)^3(2x+9)}{4x^3+3}-\frac{(x+3)^6(2x+9)^2}{2(4x^3+3)^2}+..\infty))\)

\(=lt_{x\rightarrow -3}(\frac{1}{x+3}^3\times(\frac{(x+3)^3(2x+9)}{4x^3+3}-\frac{(x+3)^6(2x+9)^2}{2(4x^3+3)^2}+..\infty))\)

\(=lt_{x\rightarrow -3}(\frac{(2x+9)}{4x^3+3}-\frac{(x+3)^3(2x+9)^2}{2(4x^3+3)^2}+…\infty)\)

All the terms except the first one go to zero, we now have

\(=lt_{x\rightarrow -3}\frac{(2x+9)}{(4x^3+3)}=\frac{2(-3)+9}{4(-3)^3+3}=\frac{3}{-105}\)

\(=-\frac{1}{35}\)