Question

Find the value of $\underset{x\to 2}{lim}\frac{x^7-128}{x^5-32}$

• A. $\frac{5}{28}$
• B. $\frac{7}{5}$
• C. $\frac{28}{5}$
• D. $\frac{5}{7}$

Answer 1

The correct option is C $\frac{28}{5}$

$L={lim}_{x\to 2}\frac{x^7-128}{x^5-32}$

This is $\frac{0}{0}$ form.

Therefore, by using L'Hospital rule:

$L={lim}_{x\to 2}\frac{{7x}^6}{{5x}^4}L={lim}_{x\to 2}\frac{7}{5}{x}^{2}L=\frac{7}{5}{\left(2\right)}^2=\frac{28}{5}$

So, option C is correct.