# If $f(x) + 3x^2 = 2f(1-x)$ and $\lim _{x\to 1}f(x) =7$, find $\lim _{x\to 0} f(x)$.

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 $$\lim_{x\to 0} f(x)=14$$.

• As for your mistake: in the last line of your attempt, plugging u into the function will not necessarily give the same value as plugging x into it does, since $u = 1 - x$ which is not equal to $x$. Sep 11, 2021 at 11:05
• @sonicsid. In the line $\lim_{u\to 0}f(u)=\lim_{x\to 0}f(x)=5$ both $u$ and $x$ are only dummy variables. The $\lim$ expressions are therefore equivalent. Sep 11, 2021 at 12:00
• @sonicsid. In both cases the argument of $f$ tends to $0$. Sep 11, 2021 at 13:22
• @md2perpe Got it now. Thank you. Apologies for causing confusion. Sep 11, 2021 at 13:24

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 $$7$$.

• so sonicsid's comment is incorrect? Sep 11, 2021 at 11:15
• Yes that comment is wrong. And the above answer is right. Sep 11, 2021 at 11:18
• @AndrewP. But their first comment (now deleted) is insightful (see the edit) Sep 11, 2021 at 11:23
• @ArcticChar After finding the function using that insight and seeing that the limit at x =1 was not to 7, it did not occur to me that such a function as stated in the question may not be possible at all and I thus thought it best to get rid of the comment altogether. Thank you for confirming my computation. Sep 11, 2021 at 13:32