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

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  • $\begingroup$ 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 $x$. $\endgroup$
    – sonicsid
    Sep 11, 2021 at 11:05
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    $\begingroup$ @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. $\endgroup$
    – md2perpe
    Sep 11, 2021 at 12:00
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    $\begingroup$ @sonicsid. In both cases the argument of $f$ tends to $0$. $\endgroup$
    – md2perpe
    Sep 11, 2021 at 13:22
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    $\begingroup$ @md2perpe Got it now. Thank you. Apologies for causing confusion. $\endgroup$
    – sonicsid
    Sep 11, 2021 at 13:24

1 Answer 1

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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 $7$.

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  • $\begingroup$ so sonicsid's comment is incorrect? $\endgroup$
    – Andrew P.
    Sep 11, 2021 at 11:15
  • $\begingroup$ Yes that comment is wrong. And the above answer is right. $\endgroup$
    – void_117
    Sep 11, 2021 at 11:18
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    $\begingroup$ @AndrewP. But their first comment (now deleted) is insightful (see the edit) $\endgroup$ Sep 11, 2021 at 11:23
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    $\begingroup$ @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. $\endgroup$
    – sonicsid
    Sep 11, 2021 at 13:32

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