Question

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

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

Try using the Taylor series expansion for sin(x) or rearrange the expression to use standard limits like (x - sin x)/x^3.

Answer 1

The Correct Option is B

We can solve this limit by rearranging terms to use standard limit forms. We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
Let the limit be $L$:
$$L = \lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x}$$
Divide the numerator and denominator by $x^4$:
$$L = \lim_{x \to 0} \frac{\frac{\sin^2 x}{x^2}}{\frac{x^2 - \sin^2 x}{x^4}}$$
Since $\lim_{x \to 0} \frac{\sin^2 x}{x^2} = 1$, the limit becomes:
$$L = \frac{1}{\lim_{x \to 0} \frac{x^2 - \sin^2 x}{x^4}}$$
Let's evaluate the denominator limit $D = \lim_{x \to 0} \frac{x^2 - \sin^2 x}{x^4}$. We can factor the numerator:
$$D = \lim_{x \to 0} \frac{(x - \sin x)(x + \sin x)}{x^4} = \lim_{x \to 0} \left( \frac{x - \sin x}{x^3} \cdot \frac{x + \sin x}{x} \right)$$
Using standard limits:
1) $\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac{1}{6}$.
2) $\lim_{x \to 0} \frac{x + \sin x}{x} = \lim_{x \to 0} (1 + \frac{\sin x}{x}) = 1 + 1 = 2$.
Therefore, $D = \frac{1}{6} \times 2 = \frac{1}{3}$.
Finally, $L = \frac{1}{D} = \frac{1}{1/3} = 3$.