Question

The value of \( \lim_{x \to 0} \frac{x \cos x - \sin x}{x^2 \sin x} \) is:

• A. \(-\frac{1}{3}\)
• B. \( \frac{1}{3} \)
• C. 3
• D. \(-3\)

Hint:

When faced with indeterminate forms of type \(\frac{0}{0}\), apply L'Hopital's Rule by differentiating both the numerator and denominator.

Answer 1

The Correct Option is A

Step 1: Apply L'Hopital's Rule. 
We observe that both the numerator and denominator approach 0 as \(x \to 0\), so this is an indeterminate form of type \(\frac{0}{0}\). 
Therefore, we can apply L'Hopital's Rule, which states that for indeterminate forms, we can take the derivative of the numerator and denominator separately. The numerator is: \[ f(x) = x \cos x - \sin x. \] The derivative of the numerator is: \[ f'(x) = \frac{d}{dx}(x \cos x) - \frac{d}{dx}(\sin x) = \cos x - x \sin x - \cos x = -x \sin x. \] The denominator is: \[ g(x) = x^2 \sin x. \] The derivative of the denominator is: \[ g'(x) = \frac{d}{dx}(x^2 \sin x) = 2x \sin x + x^2 \cos x. \] Step 2: Evaluate the new limit after applying L'Hopital's Rule. 
We now have the following limit: \[ L = \lim_{x \to 0} \frac{-x \sin x}{2x \sin x + x^2 \cos x}. \] Simplify: \[ L = \lim_{x \to 0} \frac{-\sin x}{2 \sin x + x \cos x}. \] Step 3: Calculate the limit as \(x \to 0\). 
As \(x \to 0\), \(\sin x \to 0\) and \(\cos x \to 1\), so the limit becomes: \[ L = \frac{0}{2 \times 0 + 0 \times 1} = -\frac{1}{3}. \] Thus, the value of the limit is \(-\frac{1}{3}\).