Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (c)/Solution 1

If you try to evaluate the limit directly, you end up with $\mathrm{\infty }/\mathrm{\infty }$, which means we cannot conclude anything for the moment, more work needs to be done. Here we have

$\begin{array}{rl}\underset{x\to \mathrm{\infty }}{\lim }\frac{2x}{\sqrt{9x^2+x-1}}& =\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{\frac{\sqrt{9x^2+x-1}}{x}}\\ & =\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{\frac{\sqrt{9x^2+x-1}}{\sqrt{x^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{\sqrt{\frac{9x^2+x-1}{x^2}}}\\ & =\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{\sqrt{9+\frac{1}{x}-\frac{1}{x^2}}}\\ & =\frac{2}{\sqrt{9}}\\ & =\frac{2}{3}.\end{array}$