Answer :

To find the limit \(\lim_{x \rightarrow \theta} \frac{x \sin \theta - \theta \sin x}{x - \theta}\), we can proceed by carefully analyzing and simplifying the expression step by step.

1. Define the limit expression:

[tex]\[ L = \lim_{x \rightarrow \theta} \frac{x \sin \theta - \theta \sin x}{x - \theta} \][/tex]

2. Recognize that direct substitution of \(x = \theta\) leads to an indeterminate form \( \frac{0}{0} \).

This suggests that we might need to use L'Hospital's Rule, which is applied to evaluate limits of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). L'Hospital's Rule states that:

[tex]\[ \lim_{x \rightarrow \theta} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \theta} \frac{f'(x)}{g'(x)} \][/tex]

if the limit exists.

3. Identify \(f(x)\) and \(g(x)\) from the expression:

Here,
[tex]\[ f(x) = x \sin \theta - \theta \sin x \][/tex]
[tex]\[ g(x) = x - \theta \][/tex]

4. Differentiate \(f(x)\) and \(g(x)\) with respect to \(x\):

The derivative of \(f(x)\) with respect to \(x\) is:
[tex]\[ f'(x) = \frac{d}{dx} \left( x \sin \theta - \theta \sin x \right) = \sin \theta - \theta \cos x \][/tex]

The derivative of \(g(x)\) with respect to \(x\) is:
[tex]\[ g'(x) = \frac{d}{dx} \left( x - \theta \right) = 1 \][/tex]

5. Apply L'Hospital's Rule:

[tex]\[ L = \lim_{x \rightarrow \theta} \frac{\sin \theta - \theta \cos x}{1} \][/tex]

6. Simplify the expression under the limit:

[tex]\[ L = \lim_{x \rightarrow \theta} (\sin \theta - \theta \cos x) \][/tex]

7. Evaluate the limit as \(x\) approaches \(\theta\):

By direct substitution, we have:
[tex]\[ L = \sin \theta - \theta \cos \theta \][/tex]

Therefore, the limit is:

[tex]\[ \boxed{-\theta \cos \theta + \sin \theta} \][/tex]

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