If $f(x) + 3x^2 = 2f(1-x)$ and $\lim _{x\to 1}f(x) =7$, find $\lim _{x\to 0} f(x)$.
I tried to solve this problem with this method: $$\lim_{x\to 1}f(x)= \lim_{x\to 1}2f(1-x)-\lim_{x\to 1}3x^2$$
$$7= \lim _{x\to 1}2f(1-x) - 3 \Rightarrow 5=\lim_{x\to 1}f(1-x)$$
Putting $u=1-x $
$$\lim_{x\to 1}(1-x)=0$$
$$\lim_{u\to 0}f(u)=\lim_{x\to 0}f(x)=5$$
Apperently this method is incorrect so could you point out the mistake for me? The correct answer is $\lim_{x\to 0} f(x)=14$.
Look like whoever give you the problem want you to take limit to $0$ directly:
$$ \lim_{x\to 0} f(x) + \lim_{x\to 0} 3x^2 = \lim_{x\to 0} 2f(1-x).$$
Then you obtain $\lim_{x\to 0} f(x) + 3\cdot 0 = 14$, thus the limit is $14$.
But they are not careful enough to see that your method also works. Or put it another way, there is no function $f$ that satisfies the property so that the limit at $0$ exists.
Following an idea from a (deleted) comment, we can find $f(x)$ explicitly (assuming only the functional equation, but not the limit at $1$). Put $x\mapsto 1-x$ in the equation, we have the system of equations
\begin{align} f(x) + 3x^2 &=2f(1-x),\\ f(1-x) + 3(1-x)^2 &= 2 f(x). \end{align}
Solving for $f(x)$ gives
$$f(x)+3x^2 + 6(1-x)^2 = 4f(x),$$ thus $f(x) = x^2 + 2(1-x)^2$. It is clear that for this $f$, the limit at $x=1$ is not $7$.