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I know definition of limits, continuity and so on, and I know from calculus that every polynomial is continuous in its domain but I am thinking about it too much and having trouble understanding this clearly. For example polynomial P(x) is given by $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1 x+ a_0$$ then is it really is continuous at $x = 0$? I know that continuity at $x=a$ is defined by having right and left limits which exist and are equal to each other and to the value at $x=a$, but what if all the points to the left are less than $a$ and all the points to the right are greater than $a$? For clarification, suppose you have the function $$P(x)=x^3+5$$ In this case if we take limit from right side as $x$ approaches $0$ our result must be more than $5$ because it is always above $5$ when $x>0$, right? And the left limit will always be less than $5$, so why we can say that the limit exist at $x=0$? It's left and right limits are different by one. Thanks a lot guys.

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You're question is phrased in a very confusing way. Can you try to clean it up? – Alex Becker Dec 23 '11 at 8:36
@Alex it is a just basic question which i wanted to ask,i took zero because our polynomial is defined at zero(it's domain includes zero) – dato datuashvili Dec 23 '11 at 8:41
I tried my best to clean up your question to make it understandable. Is this what you were trying to ask? – Alex Becker Dec 23 '11 at 8:54
up vote 9 down vote accepted

Yes. A polynomial over the real numbers is continuous at all points, including 0. I think your confusion is stemming from the fact that even if $f(x) < c$ for all $x < 0$, we can still have $\lim\limits_{x^-\rightarrow 0}f(x) = c$. The reason this can be true, such as $\lim\limits_{x^-\rightarrow 0}(x^3+5) = 5$ in the example you gave, is that for the limit to be $5$ we don't need points higher and lower than $5$, but merely that the values $f(x)$ for $x$ in $(-\delta,0)$ are arbitrarily close to 5 for sufficiently small $\delta>0$. It is fine if these are all less than 5, so long as how much less than 5 they are gets arbitrarily small.

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yes it is thanks @Alex – dato datuashvili Dec 23 '11 at 9:01
sorry if my question was too big – dato datuashvili Dec 23 '11 at 9:02

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