The value of _x0((x^2^2x)/(x^2-^2x)) is:

Question

The value of $\displaystyle\lim_{x	o0}\left(\frac{x^2\sin^2x}{x^2-\sin^2x}ight)$ is:

Accepted Answer

Using $\sin x=x-\dfrac{x^3}{6}+\cdots$, we get $\sin^2x=x^2-\dfrac{x^4}{3}+\cdots$, so
$$x^2-\sin^2x=\frac{x^4}{3}+\cdots$$

while the numerator $x^2\sin^2x\approx x^4$ to leading order. Hence
$$\lim_{x\to0}\frac{x^2\sin^2x}{x^2-\sin^2x}=\lim_{x\to0}\frac{x^4}{x^4/3}=3$$

Solutions · 1