Question

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

• A. $\frac{\log 3-1}{\log 3+1}$
• B. $\frac{\log 3+1}{\log 3-1}$
• C. $1$
• D. ${\log }_3\left(\frac{3}{e}\right)$

Answer 1

The correct option is A $\frac{\log 3-1}{\log 3+1}$

Let, $L=\underset{x\to 3}{lim}\frac{3^x-x^3}{x^x-3^3}=\underset{x\to 3}{lim}\frac{e^{\log 3^x}-x^3}{e^{\log x^x}-3^3}$

$=\underset{x\to 3}{lim}\frac{e^{x\log 3}-x^3}{e^{x\log x}-3^3}$

Form of the limit is $\frac{0}{0}$, thus using L-Hospital's rule, we get

$\therefore L=\underset{x\to 3}{lim}\frac{e^{x\log 3}\log 3-3x^2}{e^{x\log x}(\log x+1)}=\underset{x\to 3}{lim}\frac{3^x\log 3-3x^2}{x^x(\log x+1)}$

$=\frac{3^3\log 3-3^3}{3^3(\log 3+1)}=\frac{\log 3-1}{\log 3+1}$