3. find the following limit: lim(x→0) (sin x - x)/x³

Explanation:

Step1: Apply L'Hopital's rule

Since it is in $\frac{0}{0}$ - form when $x\rightarrow0$ (as $\sin(0)-0 = 0$ and $0^{3}=0$), we differentiate the numerator and denominator. The derivative of $\sin x - x$ is $\cos x - 1$, and the derivative of $x^{3}$ is $3x^{2}$. So the limit becomes $\lim_{x\rightarrow0}\frac{\cos x - 1}{3x^{2}}$.

Step2: Apply L'Hopital's rule again

The new - limit is still in $\frac{0}{0}$ - form (as $\cos(0)-1=1 - 1 = 0$ and $3\times0^{2}=0$). Differentiating the numerator and denominator again, the derivative of $\cos x - 1$ is $-\sin x$, and the derivative of $3x^{2}$ is $6x$. So the limit becomes $\lim_{x\rightarrow0}\frac{-\sin x}{6x}$.

Step3: Apply the well - known limit

We know that $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$. So $\lim_{x\rightarrow0}\frac{-\sin x}{6x}=-\frac{1}{6}\lim_{x\rightarrow0}\frac{\sin x}{x}=-\frac{1}{6}\times1$.

Answer:

$-\frac{1}{6}$