Question

Find the value of $\underset{x\to 1}{lim}\frac{x^x-1}{x\log x}-\underset{x\to 0}{lim}\frac{\log (1-3x)}{x}$

Answer 1

Assume $l_1=li{m}_{x\to 1}\frac{x^x-1}{x\log x}$ and $l_2=li{m}_{x\to 0}\frac{\log (1-3x)}{x}$
Applying L hospital's rule to $l_1$
$l_1=li{m}_{x\to 1}\frac{x.x^{x-1}}{\log x+1}=1$
and $l_2=li{m}_{x\to 0}\frac{\log (1-3x)}{-3x}(-3)=-3\left(1\right)=-3$
Hence $l_1-l_2=4$