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
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
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
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)$.
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?