Problem - 4597
Let $f(x)$ be a twice differentiable continuous function, and $f(0)=f'(0)=0$, $f''(0)=6$. Find the value of $$\lim_{x\to 0}\frac{f\left(\sin^2{x}\right)}{x^4}$$
Expand $f(x)$ around $x=0$ gives
$$f(x)=f(0)+f'(0)x + \frac{1}{2}f''(0)x^2 + O\left(x^3\right)=3x^2 + O\left(x^3\right)$$
Therefore,
$$f(\sin^2{x})=3\sin^4{x} + O(\sin^6{x})\implies\lim{x\to 0}\frac{f\left(\sin^2x\right)}{x^4}=\lim_{x\to 0}\frac{3\sin^4{x} + O(\sin^6{x})}{x^4}=\boxed{3}$$