Question

$\lim_{x\rightarrow0}\frac{x \cos^{2}x}{\sin x}$ is equal to ________.

• A. 4
• B. 2
• C. -2
• D. 0
• E. 1

Hint:

$\lim_{x\to 0} \frac{x}{\sin x}$ is 1.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
This is a limit problem which results in the indeterminate form \(\frac{0}{0}\) upon direct substitution of \(x=0\). We can solve it by rearranging the terms to make use of the fundamental trigonometric limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).
Step 2: Key Formula or Approach
We will use the following standard limits:
1. \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\), which implies \(\lim_{\theta \to 0} \frac{\theta}{\sin \theta} = 1\).
2. The limit of a product is the product of the limits, provided they exist.
Step 3: Detailed Explanation
1. Rearrange the expression.
The given limit is:
\[ L = \lim_{x \to 0} \frac{x\cos^2 x}{\sin x} \] We can regroup the terms to isolate the \(\frac{x}{\sin x}\) part.
\[ L = \lim_{x \to 0} \left( \frac{x}{\sin x} \cdot \cos^2 x \right) \] 2. Apply the limit product rule.
\[ L = \left( \lim_{x \to 0} \frac{x}{\sin x} \right) \cdot \left( \lim_{x \to 0} \cos^2 x \right) \] 3. Evaluate each limit separately.
- For the first limit, we use the reciprocal of the standard sine limit: \[ \lim_{x \to 0} \frac{x}{\sin x} = \lim_{x \to 0} \frac{1}{\frac{\sin x}{x}} = \frac{1}{1} = 1 \] - For the second limit, we can directly substitute \(x=0\): \[ \lim_{x \to 0} \cos^2 x = (\cos 0)^2 = (1)^2 = 1 \] 4. Multiply the results.
\[ L = 1 \cdot 1 = 1 \] Step 4: Final Answer
The value of the limit is 1.