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

$f(x)=\begin{cases}5x-4 & 0<x \leq 1\\4x^2 -3x & 1 < x < 2.\end{cases}$

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Note how I formatted the piecewise function here, for future reference. – Cameron Buie Jul 4 '12 at 14:51
I am still learning Tex – HackToHell Jul 4 '12 at 14:52
It's all good. I've been using it for years, and I'm still learning new things about it. – Cameron Buie Jul 4 '12 at 15:02
An equation is not the sort of thing that can be continuous. The question is really about whether a particulat function is continuous. It's important to pay attention to terminology becaue the terminology reflects the concepts required to address the question. – Carl Mummert Jul 4 '12 at 15:05
@CarlMummert I told you I was a beginner, anyway edited the title... – HackToHell Jul 4 '12 at 15:10
up vote 1 down vote accepted

Suppose $a,b\in\Bbb R$ with $a<b$, 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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I get it, as we move closer to the point on the graph from both sides, the net result must be zero ! must not miss classes – HackToHell Jul 4 '12 at 15:02
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. – Cameron Buie Jul 4 '12 at 15:08

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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In which equation must I check the left and right limits ? – HackToHell Jul 4 '12 at 14:42
So f(1) = 1 for both functions, so that's good. – RaistlinPicard Jul 4 '12 at 14:48
You need to check limit as x goes to 1 from left of 5X-4 =1, – RaistlinPicard Jul 4 '12 at 14:49
And, you need to check that limit as x goes to 1 from right of 4X^2-3X = 1 – RaistlinPicard Jul 4 '12 at 14:49
So I must substitute 1 in both functions because they deal with different ranges... Thanks :) – HackToHell Jul 4 '12 at 14:51

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