$f$ is a function at the interval $I$.
$f(x)=P_r(x)+h_r(x)$ for every real number $r>0$
$P_r$ is a polynomial. $|h_r(x)|\leq r$ for every $x$ at the $I$.
Show that $f$ is continuous at $I$.
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$h_r(x)$ is a polynomial or an arbitrary function? – Marios Gretsas Sep 15 '16 at 12:52
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exercise says just that $P_r$ is a polynomial – Please help Sep 15 '16 at 13:05
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$$\sup_{x\in I} |f(x) -P_{\frac{1}{n} } (x) |=\sup_{x\in I} |h_{\frac{1}{n} } (x) |\leq \frac{1}{n}$$ thus $f$ is a limit of uniformly convergent sequence of polynomials $(P_{\frac{1}{n} } )$ and therefore it is continuous.
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Let $a$$\in$$I$ and $\epsilon$$>0$. $P_r(X)$ is continuous at $I$ because it is a polynomial hence there exist $\delta>0$ such that for every $x$$\in$ $I$ such that $|x-a|<\delta$ we have $|P_r(x)-P_r(a)|<\epsilon/2$.
$|f(x)-f(a)|$$=|P_r(x)-P_r(a)+h_r(x)-h_r(a)|$$\leqslant$ $\epsilon/2 +2r$
Take $r<\epsilon/2$ and by the assumption that $P_r$ is a polynomial and continuous for every $r>0$ we have $|f(x)-f(a)|<\epsilon$
Marios Gretsas
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i fixed an $\epsilon$ and i used definition of continuity continuity to make the $|P_r(x)-P_r(a)|$ arbitrarily small.. for every epsilon the quantity must be arbitrarily small. – Marios Gretsas Sep 15 '16 at 18:21
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you can see why i took $\epsilon$/2 by understanding the quantifier $\forall$ and the frase $\forall$ $\epsilon$$>0$ – Marios Gretsas Sep 15 '16 at 18:31