Question

Evaluate the following limit.​

Answer 1

Step-by-step explanation:

We have,

$\lim_{x \rarr1} \frac{ {x}^{x} - x}{1 - x + log(x) }$

Since, it is 0/0 form so using l'hospital rule

$= \lim_{x \rarr1} \frac{ {x}^{x} ( log(x) - 1)- 1}{ - 1 + \frac{1}{x} }$

$= \lim_{x \rarr1} \frac{ {x}^{x} ( log(x) - 1)^{2} + {x}^{x} . \frac{1}{x} }{ - \frac{1}{x^{2} } }$

$= \frac{ 1(0 - 1)^{2} + 1 }{ - 1 }$

$= - 2$