Question

The value of $\underset{x\to 3}{\lim }\frac{x^5-3^5}{x^8-3^8}$ is

• A. $\frac{5}{8}$
• B. $\frac{5}{64}$
• C. $\frac{5}{216}$
• D. $\frac{1}{27}$
• E. None of these.

Answer 1

The correct option is C

$\frac{5}{216}$

Explanation of the correct option.

Compute the required value.

Given : $\underset{x\to 3}{\lim }\frac{x^5-3^5}{x^8-3^8}$

$$\begin{gathered} =\frac{3^5-3^5}{3^8-3^8} \\ =\frac{0}{0} \end{gathered}$$

Since it is $\frac{0}{0}$ form, apply L.Hospital's rule,

$$\begin{gathered} \underset{x\to 3}{\lim }\frac{5x^4}{8x^7}\left[\frac{d}{dx}{x}^n=n.x^{n-1}\right] \\ =\frac{5{\left(3\right)}^4}{8{\left(3\right)}^7} \\ =\frac{5}{8{\left(3\right)}^3} \\ =\frac{5}{216} \end{gathered}$$

Therefore, the value of $\underset{x\to 3}{\lim }\frac{x^5-3^5}{x^8-3^8}$ is $\frac{5}{216}$.

Hence,option $\left(C\right)$ is the correct option.