Certainly! To solve the limit [tex]\(\lim_{x \to 0} \frac{1 - \cos x}{x \sin x}\)[/tex], we'll go through the following steps:
1. Understand the limit expression.
We are asked to find the limit of [tex]\(\frac{1 - \cos x}{x \sin x}\)[/tex] as [tex]\(x\)[/tex] approaches 0. Notice that both the numerator and the denominator approach 0 as [tex]\(x\)[/tex] approaches 0, which is an indeterminate form [tex]\(\frac{0}{0}\)[/tex].
2. Use L'Hôpital's Rule.
To resolve the [tex]\(\frac{0}{0}\)[/tex] indeterminate form, we can use L'Hôpital's Rule, which states that if the limit [tex]\(\lim_{x \to c} \frac{f(x)}{g(x)}\)[/tex] is of the indeterminate form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], we can take the derivatives of the numerator and the denominator:
[tex]\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
\][/tex]
3. Differentiate the numerator and the denominator:
Let's differentiate the numerator [tex]\(1 - \cos x\)[/tex] and the denominator [tex]\(x \sin x\)[/tex] with respect to [tex]\(x\)[/tex]:
- The derivative of [tex]\(1 - \cos x\)[/tex] is [tex]\( \sin x \)[/tex].
- For the denominator [tex]\(x \sin x\)[/tex], we'll use the product rule: [tex]\((uv)' = u'v + uv'\)[/tex].
Here, [tex]\( u = x \)[/tex] and [tex]\( v = \sin x \)[/tex]:
[tex]\[
\frac{d}{dx} (x \sin x) = \frac{d}{dx} (x) \cdot \sin x + x \cdot \frac{d}{dx} (\sin x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x
\][/tex]
4. Apply L'Hôpital's Rule:
Now we can apply L'Hôpital's Rule:
[tex]\[
\lim_{x \to 0} \frac{1 - \cos x}{x \sin x} = \lim_{x \to 0} \frac{\sin x}{\sin x + x \cos x}
\][/tex]
5. Simplify the resulting expression:
Simplify the expression inside the limit:
[tex]\[
\lim_{x \to 0} \frac{\sin x}{\sin x + x \cos x}
\][/tex]
We can divide the numerator and the denominator by [tex]\(x\)[/tex]:
[tex]\[
\lim_{x \to 0} \frac{\frac{\sin x}{x}}{\frac{\sin x}{x} + \cos x}
\][/tex]
Use the well-known limits [tex]\(\lim_{x \to 0} \frac{\sin x}{x} = 1\)[/tex] and [tex]\(\lim_{x \to 0} \cos x = 1\)[/tex]:
[tex]\[
\lim_{x \to 0} \frac{1}{1 + 1} = \frac{1}{2}
\][/tex]
Finally, the limit is:
[tex]\[
\lim_{x \to 0} \frac{1 - \cos x}{x \sin x} = \frac{1}{2}
\][/tex]
