Question

$\underset{x\to 1}{lim}\frac{x^x-x}{1-x+\log x}=$

• A. $-\frac{1}{3}$
• B. $-1$
• C. $-2$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $-2$

$\underset{x\to 1}{lim}\frac{x^x-x}{1-x+\log x}$
It is of the form $\frac{0}{0}$. So, applying L-Hospital's rule,
$=\underset{x\to 1}{lim}\frac{x^x(1+\log x)-1}{-1+\frac{1}{x}}$
Again of the form $\frac{0}{0}$, so applying L-Hospital's rule,
$\underset{x\to 1}{lim}\frac{x^x(1+\log x)+x^x\log x(1+\log x)+x^{x-1}}{-\frac{1}{x^2}}$
$=-2$