if x approaching 3 from the left =22 and x approaching 3 from the right = 22 but the lim f(x) as x approaches 3 = 20. is the function continuous at x=3? why or why not

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if x approaching 3 from the left =22 and x approaching 3 from the right = 22 but the lim f(x) as x approaches 3 = 20. is the function continuous at x=3? why or why not

Mathematics
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? if the limit from the left is 22 and the limit from the right is 22, then the "limit' cannot be 20
im given a piecewise
f(x)= 2x^2 +4 when x<3 20 when x=3 28-2x when x>3

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Other answers:

and im asked to find the limit of x to 3+ x to 3- x to 3
piecwise or not, if \[\lim_{x\rightarrow 3^-}f(x)=22=\lim_{x \rightarrow 3^+}f(x)\] then the limit is 22, not something else
the function is not continuous at the point x=3
ah the limit is 22, but the function is not continuous at 3 because it is not the same as the value of the limit there
why is that unklerhaukus
for the function to be continuous, it must have the same value as the limit
so since \[\lim_{x\rightarrow 3} f(x)=22\] but \[f(3)=20\] it is not continuous at 3
ok
Satellite73 has already said this but The function is Only continuous at some point if the limit Exists And is Equal to the function at that point.
ok can you help me with this one state the definition of a function f(x) which is continuous at x= a
Yes, actually this (following) statement is pretty confusing "the lim f(x) as x approaches 3 = 20."
i was wrong there
A function is continuous at x=a if \[ \large \space \lim \limits_{x\rightarrow a^-}f(x)=f(a)=\lim \limits_{x \rightarrow a^+}f(x) \]
(assuming the limit exists of course)
If \( \space \lim \limits_{x\rightarrow a^-}f(x)=\lim \limits_{x \rightarrow a^+}f(x) \) then the limit exists.

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