Question

Help,understanding continuity :(
{pre-calculus- Limits}

Answer 1

so is a piecewise function
so notice that "x" is greater or lesser at 0, the limit point
so you can split both funcions of the piecewise into one-sided limit checks

if each check yields the same value, THEN that means the function is "continous" at that point, or 0

because that will mean that the function on x <0 gets to "whatever"
and that function on \(x\ge 0\) gets to "whatever"

and if that "whatever" is the same...then they both meet at that point, making the graph a continuous one
thus \(\large { lim_{x\to 0}\quad f(x)
\begin{cases}
5x-8& x<0\\
|-4-x|& x\ge 0
\end{cases}
\\ \quad \\
\implies lim_{x\to 0^{\color{blue}{ -}}}\ 5x-8\quad and\quad lim_{x\to 0^{\color{blue}{ +}}}\ |-4-x| }\)

Answer 2

I get \[f(x)=f(0)=|-4-0|=4\]

Answer 3

yeap I was thinking about the absolute value one.... and the + bit
but I think for the absolute value expression, and + part, the positive version of it is used only

Answer 4

oh ok but for 5x-8

\[5(0)-8=-8\]

they are not equal does this mean that limit does not exist

Answer 5

nope.... well

it means that the "two-sided limit" does not exist, yes
because
the two one-sided limits do not match

Answer 6

\(\large lim_{x\to 0}\quad f(x)
\begin{cases}
5x-8& x<0\\
|-4-x|& x\ge 0
\end{cases}\) is asking for the two-sided limit

Answer 7

ohhhhhh ok, so did I do it correctly, like what I did, plugging in for x?

Answer 8

oooh ok thanks:)

Answer 9

yw