Question

$\underset{x\to a}{lim}\frac{x^{\frac{5}{8}}-a^{\frac{5}{8}}}{x^{\frac{1}{3}}-a^{\frac{1}{3}}}=$

• A. $\frac{15}{8}{a}^{\frac{24}{7}}$
• B. $\frac{8}{15}{a}^{\frac{24}{7}}$
• C. $\frac{15}{8}{a}^{\frac{7}{24}}$
• D. $\frac{8}{15}{a}^{\frac{7}{24}}$

Answer 1

The correct option is C $\frac{15}{8}{a}^{\frac{7}{24}}$

$\underset{x\to a}{lim}\frac{x^{\frac{5}{8}}-a^{\frac{5}{8}}}{x^{\frac{1}{3}}-a^{\frac{1}{3}}}=\frac{a^{\frac{5}{8}}-a^{\frac{5}{8}}}{a^{\frac{1}{3}}-a^{\frac{1}{3}}}=\frac{0}{0}$

Above limit is in $\frac{0}{0}$ form.

So, differentiate both numerator and denominator using L'Hospital's rule

$\therefore \underset{x\to a}{lim}\frac{x^{\frac{5}{8}}-a^{\frac{5}{8}}}{x^{\frac{1}{3}}-a^{\frac{1}{3}}}$

$=\underset{x\to a}{lim}\frac{{\frac{5}{8}\times x}^{\frac{5}{8}-1}}{{\frac{1}{3}\times x}^{\frac{1}{3}-1}}$

$=\underset{x\to a}{lim}\frac{15}{8}{x}^{\frac{5}{8}-\frac{1}{3}}$

$=\underset{x\to a}{lim}\frac{15}{8}{x}^{\frac{7}{24}}$

$=\frac{15}{8}{a}^{\frac{7}{24}}$