Q. 55 - Calculus

Question

Calculate each limit in Exercises 35–80.
$\underset{x\to 0^{+}}{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 0

Step 1. Given information

The given expression is  $\underset{x\to 0^{+}}{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 by using the L'Hopital rule, 

$$\begin{gathered} \underset{x\to 0^{+}}{lim}\frac{x^{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-x^{{\displaystyle \raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}{x^2}=\underset{x\to 0^{+}}{lim}\left(\frac{{\displaystyle \frac{7}{2}}{x}^{{\displaystyle \frac{5}{2}}}-{\displaystyle \frac{8}{3}}{x}^{{\displaystyle \frac{5}{3}}}}{2x}\right) (\mathrm{differentiating} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator}) \\ =\underset{x\to 0^{+}}{lim}\left(\frac{{\displaystyle \frac{35}{4}}{x}^{\frac{3}{2}}-{\displaystyle \frac{45}{9}}{x}^{{\displaystyle \frac{2}{3}}}}{2}\right) (\mathrm{differentiating} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator}) \\ =\left(\frac{{\displaystyle \frac{35}{4}{\left(0\right)}^{\frac{3}{2}}-\frac{45}{9}{\left(0\right)}^{\frac{2}{3}}}}{2}\right) \\ =\frac{0-0}{2} \\ =0 \end{gathered}$$