Question

Find:
$\underset{x\to 0}{lim}\frac{\sin x-x+\frac{1}{6}{x}^3}{x^3}.$

Answer 1

Solution -

${lim}_{x\to 0}\frac{sinx-x+\frac{1}{6}{x}^3}{x^3}$

Apply L-Hospital Rule 3 times

${lim}_{x\to 0}\frac{cosx-1+\frac{1}{2}{x}^2}{3x^2}$

${lim}_{x\to 0}\frac{-sinx+x}{6x}$

${lim}_{x\to 0}\frac{-cosx+1}{6}$

$=\frac{-1+1}{6}$

$=0$