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

Solution:

we have, 

$\underset{x\to 3}{lim} \frac{3^x-x^3}{x^x-3^2}$ 

$=\underset{x\to 3}{lim} \frac{3^x{\log }_{e}3-3x^2}{x^x\left(1+{\log }_{e}x\right)-0}$   [Using L' Hospital's Rule] 

$=\frac{3^3{\log }_{e}3-3^3}{3^3\left(1+{\log }_{e}3\right)}=\frac{{\log }_{e}3-1}{{\log }_{e}3+1}$