Question

Evaluate: $\underset{x\to 1}{lim}\frac{x^2-x\log x+\mathrm{logx}-1}{x-1}$

Answer 1

$\underset{x\to 1}{lim}\frac{x^2-x\log x+\log x-1}{x-1}$

Let : $f\left(x\right)=x^2-x\log x+\log x-1$

$f^{\prime }\left(x\right)=2x-[\log x+\frac{x}{x}]+\frac{1}{x}-1$

$f^{\prime }\left(x\right)=2x-\log x+\frac{1}{x}$

$g\left(x\right)=x-1$

$g^{\prime }\left(x\right)=1$

$\underset{x\to 1}{lim}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$ [L' Hospital rule]

$=\underset{x\to 1}{lim}\frac{2x-\log x+\frac{1}{x}}{1}$

$=2-\log 1+1$

$=2+1$

$=3$