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?

• 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.
– mfl
Dec 13, 2020 at 21:43
• @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 $$\lim_{x\rightarrow 2017}2017=\lim_{x\rightarrow 2017}(2018x^2-x^3)\lim_{x\rightarrow 2017}f(x)$$ Dec 13, 2020 at 21:47

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.