Question

$\underset{x\to 2}{lim}\frac{x^5\sqrt{x}-32\sqrt{2}}{x^3\sqrt{x}-8\sqrt{2}}$=

• A. $\frac{22}{7}$
• B. $\frac{44}{7}$
• C. $6$
• D. $2$

Answer 1

The correct option is B $\frac{44}{7}$

Given limit is

$\underset{x\to 2}{lim}\frac{x^5\sqrt{x}-32\sqrt{2}}{x^3\sqrt{x}-8\sqrt{2}}$

$\Rightarrow \underset{x\to 2}{lim}\frac{x^5\sqrt{x}-2^5\sqrt{2}}{x^3\sqrt{x}-2^3\sqrt{2}}$

$\Rightarrow \underset{x\to 2}{lim}\frac{x^{\frac{11}{2}}-2^5\sqrt{2}}{x^{\frac{7}{2}}-2^3\sqrt{2}}$

Applying L' hospital Rule , we get

$\Rightarrow \underset{x\to 2}{lim}\frac{\frac{11}{2}{x}^{\frac{11}{2}-1}-0}{\frac{7}{2}{x}^{\frac{7}{2}-1}-0}$

$\Rightarrow \underset{x\to 2}{lim}\frac{11x^{4.5}}{7x^{2.5}}$

$\Rightarrow \underset{x\to 2}{lim}\frac{11x^2}{7}=\frac{11}{7}{2}^2=\frac{44}{7}$