Question

Find the limits of $\frac{1-x+\log x}{1-\sqrt{2x-x^2}},$ when $x=1$.

Answer 1

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

Using L'Hopital's Rule,

$={lim}_{x\to 1}\frac{-1+\frac{1}{x}}{\frac{2-2x}{2\sqrt{2x-x^2}}}$

Using L'Hopital's Rule,

$={lim}_{x\to 1}\frac{\sqrt{2x-x^2}}{x}=1$