Question

If $L=\underset{x\to 0}{lim}\frac{x^2{\sin }^{2}x}{x^2-{\sin }^{2}x}$ exists, then $L$ is equal to

• A. 0
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (B)

3

$L=\underset{x\to 0}{lim}\frac{x^2{\sin }^{2}x}{x^2-{\sin }^{2}x}=\underset{x\to 0}{lim}\frac{x^4\frac{{\sin }^{2}x}{x^2}}{x^2-{\sin }^{2}x}$

$\underset{x\to 0}{lim}\frac{x^4}{x^2-{\sin }^{2}x}=\underset{x\to 0}{lim}\frac{x^4}{x^2-{(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots )}^2}$

$\underset{x\to 0}{lim}\frac{x^4}{x^2-(x^2-\frac{2x^4}{3!}+\cdots higherpowersofx)}$

$\underset{x\to 0}{lim}\frac{x^4}{\frac{2x^4}{3!}+\cdots higherpowersofx}=\frac{1}{\frac{2}{3!}}=3$