Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (a)/Solution 1

Following the hint, we have

$\begin{array}{rl}& \underset{x\to -\mathrm{\infty }}{\lim }\frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}\\ & =\underset{x\to -\mathrm{\infty }}{\lim }\frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}\cdot \frac{\sqrt{x^4+x^3+2}+x^2}{\sqrt{x^4+x^3+2}+x^2}\\ & =\underset{x\to -\mathrm{\infty }}{\lim }\frac{x^4+x^3+2-x^4}{(5x+3)\cdot (\sqrt{x^4+x^3+2}+x^2)}\\ & =\underset{x\to -\mathrm{\infty }}{\lim }\frac{x^3+2}{(5x+3)(\sqrt{x^4+x^3+2}+x^2)}.\end{array}$

Now if we divide both the denominator and the numerator by $x^3$, we get

$\begin{array}{rl}& \underset{x\to -\mathrm{\infty }}{\lim }\frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}\\ & =\underset{x\to -\mathrm{\infty }}{\lim }\frac{\frac{x^3+2}{x^3}}{\frac{5x+3}{x}\cdot \frac{\sqrt{x^4+x^3+2}+x^2}{x^2}}\\ & =\underset{x\to -\mathrm{\infty }}{\lim }\frac{1+\frac{2}{x^3}}{(5+\frac{3}{x})\cdot (\sqrt{\frac{x^4+x^3+2}{x^4}}+\frac{x^2}{x^2})}\\ & =\underset{x\to -\mathrm{\infty }}{\lim }\frac{1+\frac{2}{x^3}}{(5+\frac{3}{x})\cdot (\sqrt{1+\frac{1}{x}+\frac{2}{x^4}}+1)}\\ & =\frac{1}{5\cdot (1+1)}\\ & =\frac{1}{10},\end{array}$

so $\underset{x\to -\mathrm{\infty }}{\lim }\frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}=\frac{1}{10}$.