Question

The value of $\underset{x\to \infty }{\lim }\left(\frac{x^2+bx+4}{x^2+ax+5}\right)$ is

• A. $\frac{b}{a}$
• B. $0$
• C. $1$
• D. $\frac{4}{5}$

Answer 1

The correct option is C

$1$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to \infty }{\lim }\left(\frac{x^2+bx+4}{x^2+ax+5}\right)$

Divide both numerator and denominator by the highest power of $x$,

$$\begin{gathered} \Rightarrow \underset{x\to \infty }{\lim }\left(\frac{{\displaystyle \frac{x^2}{x^2}}+{\displaystyle \frac{bx}{x^2}}+{\displaystyle \frac{4}{x^2}}}{{\displaystyle \frac{x^2}{x^2}}+{\displaystyle \frac{ax}{x^2}}+{\displaystyle \frac{5}{x^2}}}\right) \\ \Rightarrow \underset{x\to \infty }{\lim }\left(\frac{1+{\displaystyle \frac{b}{x}}+{\displaystyle \frac{4}{x^2}}}{1+{\displaystyle \frac{a}{x}}+{\displaystyle \frac{5}{x^2}}}\right) \\ \Rightarrow \frac{1+0+0}{1+0+0} \\ \Rightarrow 1 \end{gathered}$$

Therefore the value of $\underset{x\to \infty }{\lim }\left(\frac{x^2+bx+4}{x^2+ax+5}\right)$ is $1$.

Hence,option $\left(C\right)$ is the correct option.