Find the values of $a$ and $b$ such that $\lim_{x \to 0} \frac{a\sin^2x \times b\log \cos x}{x^4} = \frac{1}{2}$.
As-
$$\lim_{x \to 0} \frac{a\sin^2x \times b\log \cos x}{x^4} = \frac{0}{0}\ \text{form}$$ Note that $$\sin^2x = x^2\bigg(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots\bigg)^2,\qquad \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots$$ As$$\log(1-y)= -\bigg(y+\frac{y^2}{2}+\frac{y^3}{3}+\cdots\bigg) $$ $$\implies \log \cos x = -\bigg[\bigg(\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\bigg)+\frac{1}{2}\bigg(\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\bigg)^2+\cdots\bigg] $$ Therefore $$\lim_{x \to 0} \frac{a\sin^2x \times b\log \cos x}{x^4}\hspace{90mm}$$ $$ = \lim_{x \to 0} \frac{x^4 a\big(1-\frac{x^2}{3!}+\frac{x^4}{5!}+\cdots\big)^2 \times (-b)\big[\big(\frac{1}{2!}+\frac{x^2}{4!}+\cdots\big)+\frac{1}{2}\big(\frac{x}{2!}+\frac{x^3}{4!}+\cdots\big)^2+\cdots\big] }{x^4}$$ $$ = (-ab)\frac{1}{2} = \frac{1}{2} \implies ab = -1$$ But i need to find a and b? How to get that?