Q. 56 - Calculus

Question

Calculate each limit in Exercises 35–80. 

$\underset{x\to \infty }{lim}\frac{x^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{x^2}$

Step-by-Step Solution

Answer

The limit is infinity.

Step 1. Given information

The given expression is $\underset{x\to \infty }{lim}\frac{x^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{x^2}$

Step 2. Calculation

The limit is calculated as below using L'Hopital Rule, 

$$\begin{gathered} \underset{x\to \infty }{lim}\frac{x^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{x^2}=\underset{x\to \infty }{lim}\left(\frac{{\displaystyle \frac{7}{2}{x}^{\frac{5}{2}}-\frac{8}{3}{x}^{\frac{5}{3}}}}{2x}\right) (\mathrm{differentiating} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator}) \\ =\underset{x\to \infty }{lim}\left(\frac{{\displaystyle \frac{35}{4}{x}^{\frac{3}{2}}-\frac{45}{9}{x}^{\frac{2}{3}}}}{2}\right) (\mathrm{differentiating} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator}) \\ =\left(\frac{{\displaystyle \frac{35}{4}{(\infty )}^{\frac{3}{2}}-\frac{45}{9}{(\infty )}^{\frac{2}{3}}}}{2}\right) \\ =\frac{\infty -\infty }{2} \\ =\infty \end{gathered}$$