Question

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

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

Answer 1

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

$\underset{x\to 2}{lim}\frac{x^7-128}{x^5-32}=\underset{x\to 2}{lim}\frac{x^7-2^7}{x^5-2^5}=\underset{x\to 2}{lim}(\frac{x^7-2^7}{x-2}.\frac{x-2}{x^5-2^5})=\underset{x\to 2}{lim}(\frac{x^7-2^7}{x-2}÷\frac{x^5-2^5}{x-2})=\underset{x\to 2}{lim}\frac{x^7-2^7}{x-2}÷\underset{x\to 2}{lim}\frac{x^5-2^5}{x-2}=7.2^{7-1}÷5.2^{5-1}=\frac{7}{5}{2}^{7-5}=\frac{7}{5}{2}^2=\frac{28}{5}$

Hence, option $C$ is the correct answer.