Question

$\underset{x\to \infty }{\lim }\sqrt{x^2+7x-x}$

Answer 1

$$\begin{gathered} \underset{x\to \infty }{\lim }\left[\sqrt{x^2+7x}-x\right] \\ \mathrm{Rationalising}\mathrm{the}\mathrm{numerator}: \\ \underset{x\to \infty }{\lim }\left[\left(\sqrt{x^2+7x}-x\right)\frac{\left(\sqrt{x^2+7x}+x\right)}{\left(\sqrt{x^2+7x}+x\right)}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{x^2+7x-x^2}{\left(\sqrt{x^2+7x}+x\right)}\right] \\ \mathrm{Dividing}\mathrm{the}\mathrm{numerator}\mathrm{and}\mathrm{the}\mathrm{denominator}\mathrm{by}x: \\ \underset{x\to \infty }{\lim }\left[\frac{7}{{\displaystyle \frac{\sqrt{x^2+7x}}{x}}+1}\right] \\ =\underset{x\to \infty }{\lim }\left[\frac{7}{{\displaystyle \frac{\sqrt{x^2+7x}}{x}}+1}\right] \\ \mathrm{When}x\to \infty ,\mathrm{then}\frac{1}{x}\to 0. \\ \Rightarrow \frac{7}{\sqrt{1}+1} \\ =\frac{7}{2} \end{gathered}$$