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

To evaluate this limit, we can use the Taylor series expansion of $\sin x$ around $x = 0$. The expansion is given by:
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + O(x^5)$$
We need the expression for $\sin^2 x$:
$$\sin^2 x = \left( x - \frac{x^3}{6} + \dots \right)^2 = x^2 - 2(x)(\frac{x^3}{6}) + O(x^6) = x^2 - \frac{x^4}{3} + O(x^6)$$
Now substitute this into the given limit expression:
$$\lim_{x \to 0} \frac{x^2 \sin^2 x}{x^2 - \sin^2 x} = \lim_{x \to 0} \frac{x^2 (x^2 - \frac{x^4}{3} + \dots)}{x^2 - (x^2 - \frac{x^4}{3} + \dots)}$$
$$\lim_{x \to 0} \frac{x^4 - \frac{x^6}{3} + \dots}{\frac{x^4}{3} - \dots}$$
Dividing both numerator and denominator by $x^4$:
$$\lim_{x \to 0} \frac{1 - \frac{x^2}{3} + \dots}{\frac{1}{3} - \dots} = \frac{1}{1/3} = 3$$
Note: Option 2 corresponds to the value 3.