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Suppose $\;f(x)\;$ is a continuous function, and that $$\lim_{x\rightarrow 2017}\big(x+f(x)\big)=2018$$

I want to know if $\;y=2017\;$ intersects the graph of function $\;y=\left(2018x^2-x^3\right)f(x)\;$ multiple times.

Is it correct to say that $\;f(x)=2018-x\;$, since if we separate the limit into two limits, we get $$\lim_{x\rightarrow 2017}\big(x+f(x)\big)=\lim_{x\rightarrow2017}x+\lim_{x\rightarrow2017}f(x)=2018\Rightarrow \lim_{x\rightarrow 2017}f(x)=1,$$

and because of this the limit

$$\lim_{x\rightarrow n}x=n$$

Is this correct?

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    $\begingroup$ You can say that $\lim_{x\rightarrow 2017}f(x)=1.$ And $2018-x$ verifies that condition but it is not the unique function doing it. $\endgroup$
    – mfl
    Dec 13, 2020 at 21:43
  • $\begingroup$ @mfl So essentially that would be the incorrect way to go about finding how many times the function intersects y=2017? Or is it appropriate to use ANY real function that verifies that condition in this way? Or could I say $$\lim_{x\rightarrow 2017}2017=\lim_{x\rightarrow 2017}(2018x^2-x^3)\lim_{x\rightarrow 2017}f(x)$$ $\endgroup$ Dec 13, 2020 at 21:47

1 Answer 1

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As you said, $\lim_{x\to2017}f(x)=1$ and since $f$ is continuous, $f(2017)=1$. This is all we can say about $f$ which is not enough information to determine the number of intersection points or what $f$ actually is. $f(x)=2018-x$ is only one possibility and there's infinitely many.

However, we can guarantee at least $2$ intersection points. If you let $g(x)=(2018x^2-x^3)f(x)$, then $g(0)=0, g(2017)=2017^2$, and $g(2018)=0$. $g$ is continuous (since $f$ is continuous) and by the Intermediate Value Theorem there are numbers $a\in(0,2017)$ and $b\in(2017,2018)$ such that $g(a)=2017, g(b)=2017$. So $2$ intersection points are $(a,2017)$ and $(b,2017)$.

An example with exactly $2$ intersection points is $f(x)=x/2017$. $f(x)=1$ would give you exactly $3$ intersection points. And there's an example with infinitely many intersection points.

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  • $\begingroup$ I completely understand now. This makes perfect sense. Thank you. I see now that continuity is vital in this reasoning. $\endgroup$ Dec 13, 2020 at 22:26

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