Question

use lhôpitals rule to find the following limit.

$$\lim_{x \to 0} \frac{\sin(x) - x}{3x^{3}}$$

$$\lim_{x \to 0} \frac{\sin(x) - x}{3x^{3}} = \square$$ (type an exact answer.)

Explanation:

Step1: Check the form of the limit

When \(x = 0\), \(\sin(0)-0=0\) and \(3\times0^{3}=0\). So, it is in the \(\frac{0}{0}\) form. Apply L'Hôpital's Rule.
Differentiate the numerator and denominator.
The derivative of \(y=\sin(x)-x\) is \(y'=\cos(x) - 1\), and the derivative of \(y = 3x^{3}\) is \(y'=9x^{2}\).
So, \(\lim_{x
ightarrow0}\frac{\sin(x)-x}{3x^{3}}=\lim_{x
ightarrow0}\frac{\cos(x)-1}{9x^{2}}\)

Step2: Re - check the form and apply L'Hôpital's Rule again

When \(x = 0\), \(\cos(0)-1=0\) and \(9\times0^{2}=0\). It is still in the \(\frac{0}{0}\) form.
Differentiate the numerator and denominator again.
The derivative of \(y=\cos(x)-1\) is \(y'=-\sin(x)\), and the derivative of \(y = 9x^{2}\) is \(y' = 18x\).
So, \(\lim_{x
ightarrow0}\frac{\cos(x)-1}{9x^{2}}=\lim_{x
ightarrow0}\frac{-\sin(x)}{18x}\)

Step3: Re - check the form and apply L'Hôpital's Rule again

When \(x = 0\), \(-\sin(0)=0\) and \(18\times0=0\). It is still in the \(\frac{0}{0}\) form.
Differentiate the numerator and denominator again.
The derivative of \(y=-\sin(x)\) is \(y'=-\cos(x)\), and the derivative of \(y = 18x\) is \(y'=18\).
So, \(\lim_{x
ightarrow0}\frac{-\sin(x)}{18x}=\lim_{x
ightarrow0}\frac{-\cos(x)}{18}\)

Step4: Evaluate the limit

Substitute \(x = 0\) into \(\frac{-\cos(x)}{18}\).
When \(x = 0\), \(\cos(0)=1\). So, \(\lim_{x
ightarrow0}\frac{-\cos(x)}{18}=\frac{-1}{18}\)

Answer:

\(-\frac{1}{18}\)