Let $f(x)$ be a continuous function in $\mathbb{R}$ such that $f''(1)=2,f(1)=f'(1)=0$. Prove that exists $g(x)$ continuous in $\mathbb{R}$ such that $f(x)=(x-1)^2g(x)$
The first thing I thought when I saw this is Taylor Series, around $x=1$. I know that(using Lagrange residue):
$f(x)=f(1)+f'(2)(x-1)+\frac{1}{2}f''(1)(x-1)^2+\frac{f''''(c)}{24}(x-1)^4=(x-1)^2+\frac{f''''(c)}{24}(x-1)^4=(x-1)^2[1+g(x)]$
where $g(x)=\frac{f''''(c)}{24}(x-1)^2$, and $0 \lt c \lt 1$.
Is that correct? (It is obvious that $g$ is continuous).
No, it is not correct.
L'Hospital twice gives
$$ \lim_{x \to 1} \frac{f(x)}{(x-1)^2}=1.$$
Hence define $g$ by
$g(x)=\frac{f(x)}{(x-1)^2},$ if $ x \ne 1$ and $g(1)=1.$
$g$ will do the job !