Question

Solve:
$\underset{x\to 0}{lim}\frac{\sin x-x+\frac{1}{6}{x}^3}{x^5}$

Answer 1

${lim}_{x\to 0}\frac{sinx-x+\frac{x^3}{6}}{x^5}$ this is of the form 0/0

$\Rightarrow {lim}_{x\to 0}\frac{sinx-x+\frac{x^3}{6}}{x^5}=\frac{\frac{d}{dx}(sinx-x+\frac{x^3}{6})}{\frac{d}{dx}\left(x^5\right)}$

$\Rightarrow {lim}_{x\to 0}\frac{cosx-1+\frac{3x^2}{6}}{5x^4}$

$\Rightarrow {lim}_{x\to 0}\frac{(cosx-1)+\frac{x^2}{2}}{5\left(x^4\right)}\Rightarrow \frac{0}{0}$ (Lhospital rule)

$\Rightarrow {lim}_{x\to 0}\frac{-sinx+2\left(x/2\right)}{20\left(x^3\right)}\Rightarrow \frac{0}{0}$ (L hospital rule)

${lim}_{x\to 0}\frac{-cos+1}{60x^2}\Rightarrow {lim}_{x\to 0}\frac{2si{n}^{2}x/2}{60x^2}$

${lim}_{x\to 0}\frac{si{n}^{2}x/2}{4\times x^2/4}\times \frac{2}{60}\Rightarrow 1\times \frac{2}{4\times 60}\Rightarrow \frac{1}{120}$

${lim}_{x\to 0}\frac{sinx-x+\frac{x^3}{6}}{x^5}=\frac{1}{120}$