To solve the limit [tex]\(\lim _{x \rightarrow 0} \frac{x-\sin x}{x^3}\)[/tex], let's follow a detailed, step-by-step approach.
1. Understand the function and the limit:
We need to find the limit of [tex]\(\frac{x - \sin x}{x^3}\)[/tex] as [tex]\(x\)[/tex] approaches 0.
2. Apply L'Hôpital's Rule:
L'Hôpital's Rule is applicable here because directly substituting [tex]\(x = 0\)[/tex] into the function gives an indeterminate form [tex]\(\frac{0 - 0}{0^3}\)[/tex] or [tex]\(\frac{0}{0}\)[/tex]. According to L'Hôpital's Rule, if we have an indeterminate form, we can take the derivatives of the numerator and the denominator and then re-evaluate the limit.
So we need to find the derivatives of the numerator [tex]\(x - \sin x\)[/tex] and the denominator [tex]\(x^3\)[/tex].
3. First derivative:
- Derivative of the numerator [tex]\(x - \sin x\)[/tex]:
[tex]\[
\frac{d}{dx}(x - \sin x) = 1 - \cos x
\][/tex]
- Derivative of the denominator [tex]\(x^3\)[/tex]:
[tex]\[
\frac{d}{dx}(x^3) = 3x^2
\][/tex]
Now, we apply L'Hôpital's Rule:
[tex]\[
\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{1 - \cos x}{3x^2}
\][/tex]
4. Re-evaluate the limit:
We still have an indeterminate form [tex]\(\frac{1 - 1}{0}\)[/tex]. Thus, we apply L'Hôpital's Rule a second time by differentiating the numerator and denominator again.
4. Second derivative:
- Derivative of the numerator [tex]\(1 - \cos x\)[/tex]:
[tex]\[
\frac{d}{dx}(1 - \cos x) = \sin x
\][/tex]
- Derivative of the denominator [tex]\(3x^2\)[/tex]:
[tex]\[
\frac{d}{dx}(3x^2) = 6x
\][/tex]
Applying L'Hôpital's Rule again:
[tex]\[
\lim_{x \to 0} \frac{1 - \cos x}{3x^2} = \lim_{x \to 0} \frac{\sin x}{6x}
\][/tex]
5. Simplify and evaluate:
We can simplify the limit:
[tex]\[
\lim_{x \to 0} \frac{\sin x}{6x} = \frac{1}{6} \lim_{x \to 0} \frac{\sin x}{x}
\][/tex]
We know from standard limit results that:
[tex]\[
\lim_{x \to 0} \frac{\sin x}{x} = 1
\][/tex]
Therefore:
[tex]\[
\frac{1}{6} \lim_{x \to 0} \frac{\sin x}{x} = \frac{1}{6} \times 1 = \frac{1}{6}
\][/tex]
So, the limit is:
[tex]\[
\lim _{x \rightarrow 0} \frac{x-\sin x}{x^3} = \frac{1}{6}
\][/tex]
