Question

If \(\displaystyle\lim_{x\rightarrow\infty}\left(\dfrac{x^2+5}{2x+3}-ax-b\right)=\infty\), then the value of a and b are:
1. \(a=\dfrac{1}{2}\), b ∉ ℝ
2. \(a\ne\dfrac{1}{2}\), b ∈ ℝ
3. \(a=\dfrac{1}{2}\), b ∈ ℝ
4. \(a\ne\dfrac{1}{2}\), b ∉ ℝ

Answer 1

Correct Answer - Option 2 : \(a\ne\dfrac{1}{2}\), b ∈ ℝ

Calculation:

Given:

\(\displaystyle\lim_{x\rightarrow\infty}\left(\dfrac{x^2+5}{2x+3}-ax-b\right)=\infty\)

⇒ \(\displaystyle\lim_{x\rightarrow\infty}\left(\dfrac{x^2+5-(ax+b)(2x+3)}{2x+3}\right)=∞ \) 

⇒ \(\displaystyle\lim_{x\rightarrow\infty}\left(\dfrac{x^2+5-2ax^2-2xb-3ax-3b}{2x+3}\right)=∞ \)

⇒ \(\displaystyle\lim_{x\rightarrow\infty}\left(\dfrac{x^2+5-(ax+b)(2x+3)}{2x+3}\right)=∞ \)

⇒ \(\displaystyle\lim_{x\rightarrow\infty}[(1-2a)x^2-(2b+3a)x-3b+5]=∞ \)

It is true of the degree of the numerator > degree of the denominator 

⇒ 1-2a ≠ 0 

⇒ a ≠ R - {-1/2)

⇒ b € R

⇒ \(a\ne\dfrac{1}{2}\) , b ∈ R