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I am doing basic calculus, can someone tell me how to find the continuity here at $x=1$

$f(x)=\begin{cases}5x-4 & 0

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    @CarlMummert I told you I was a *beginner*, anyway edited the title...2012-07-04

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The function is continuos because it is both right continuous and left continuos at f(1) where x=1 The limit 5x-4 =1 As x tends to 1 The limit 4x²-3x =1 As x tends to 1 So the function is both left continuous and right continuous hence generally continuous at x=1

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It's clear that both functions are continuous separately over $\Bbb R$: the problem is continuity at $x=1$. So just make sure that the limits from the right and left of $1$ of$ f(x) = f(1)$.

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    So I must substitute 1 in both functions because they deal with different ranges... Thanks :)2012-07-04
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Suppose $a,b\in\Bbb R$ with $a, let $I=(a,b)$, $c\in I$, $f:I\to\Bbb R$. Then we say that $f$ is continuous at $c$ iff $f(c)=\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x),$ that is, iff the left and right limits exist, and are both equal to $f(c)$.

In this case, coming toward $c=1$ from the left, what does $f(x)$ look like--that is, how is $f(x)$ defined for $x<1$? What about from the right?

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    Well, not *zero*, but the same. In this case, all three are equal to $1$, so it is indeed continuous at that point. Not missing classes is certainly a good plan.2012-07-04