Question

Write the value of $\underset{x\to -\infty }{\lim }\left(3x+\sqrt{9x^2-x}\right).$

Answer 1

$$\begin{gathered} \underset{x\to -\infty }{\lim }\left(3x+\sqrt{9x^2-x}\right) \\ \mathrm{Let} \\ m=-x \\ \mathrm{If}x\to -\infty ,\mathrm{then}m\to \infty . \\ =\underset{m\to \infty }{\lim }\left(-3m+\sqrt{9m^2+m}\right) \\ \mathrm{Rationalising}\mathrm{the}\mathrm{numerator},\mathrm{we}\mathrm{get}: \\ =\underset{m\to \infty }{\lim }\frac{\left(-3m+\sqrt{9m^2+m}\right)\left(-3m-\sqrt{9m^2+m}\right)}{-3m-\sqrt{9m^2+m}} \\ =\underset{m\to \infty }{\lim }\left(\frac{-\left[-9m^2+\left(9m^2+m\right)\right]}{-3m-\sqrt{9m^2+m}}\right) \\ =\underset{m\to \infty }{\lim }\left(\frac{-m}{-3m-\sqrt{9m^2+m}}\right) \\ \text{Dividing the numerator and the denominator by}m\text{, and applying limit, we get:} \\ =\frac{-1}{-3-\sqrt{9+{\displaystyle \frac{1}{\infty }}}} \\ =\frac{1}{6} \end{gathered}$$