Question

$\lim_{x\rightarrow0}\frac{\sin(\pi \sin^{2}x)}{x^{2}} = $ ________.

• A. $\frac{\pi}{2}$
• B. $\pi$
• C. $2\pi$
• D. $\pi^{2}$
• E. $\frac{\pi^{2}}{2}$

Hint:

Multiply and divide by the inner function to match the standard limit form.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This is a limit problem that results in the indeterminate form \(\frac{0}{0}\) when \(x=0\) is substituted. This suggests using the standard trigonometric limit \(\lim_{u \to 0} \frac{\sin u}{u} = 1\).
Step 2: Key Formula or Approach
We want to manipulate the expression to fit the form \(\frac{\sin u}{u}\).
1. Let \(u = \pi \sin^2 x\). As \(x \to 0\), \(\sin^2 x \to 0\), and therefore \(u \to 0\).
2. We will multiply and divide the expression by \(u\) to create the desired form.
3. We will also use the limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).
Step 3: Detailed Explanation
1. Rewrite the expression.
The limit is \(L = \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{x^2}\).
To use the standard limit, we need the argument of the sine function in the denominator. Let's multiply and divide by \(\pi \sin^2 x\).
\[ L = \lim_{x \to 0} \left[ \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \cdot \frac{\pi \sin^2 x}{x^2} \right] \] 2. Separate the limits.
Since the limits of the individual parts exist, we can separate them:
\[ L = \left( \lim_{x \to 0} \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \right) \cdot \left( \lim_{x \to 0} \frac{\pi \sin^2 x}{x^2} \right) \] 3. Evaluate the first limit.
Let \(u = \pi \sin^2 x\). As \(x \to 0\), \(u \to 0\). So the first limit becomes:
\[ \lim_{u \to 0} \frac{\sin u}{u} = 1 \] 4. Evaluate the second limit.
\[ \lim_{x \to 0} \frac{\pi \sin^2 x}{x^2} = \pi \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2 \] Using the property that \(\lim [f(x)]^n = [\lim f(x)]^n\):
\[ = \pi \left( \lim_{x \to 0} \frac{\sin x}{x} \right)^2 = \pi (1)^2 = \pi \] 5. Combine the results.
\[ L = (1) \cdot (\pi) = \pi \] Step 4: Final Answer
The value of the limit is \(\pi\).