## anonymous 4 years ago 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

1. anonymous

? if the limit from the left is 22 and the limit from the right is 22, then the "limit' cannot be 20

2. anonymous

im given a piecewise

3. anonymous

f(x)= 2x^2 +4 when x<3 20 when x=3 28-2x when x>3

4. anonymous

and im asked to find the limit of x to 3+ x to 3- x to 3

5. anonymous

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

6. UnkleRhaukus

the function is not continuous at the point x=3

7. anonymous

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

8. anonymous

why is that unklerhaukus

9. anonymous

for the function to be continuous, it must have the same value as the limit

10. anonymous

so since $\lim_{x\rightarrow 3} f(x)=22$ but $f(3)=20$ it is not continuous at 3

11. anonymous

ok

12. UnkleRhaukus

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.

13. anonymous

ok can you help me with this one state the definition of a function f(x) which is continuous at x= a

14. anonymous

Yes, actually this (following) statement is pretty confusing "the lim f(x) as x approaches 3 = 20."

15. anonymous

i was wrong there

16. anonymous

A function is continuous at x=a if $\large \space \lim \limits_{x\rightarrow a^-}f(x)=f(a)=\lim \limits_{x \rightarrow a^+}f(x)$

17. UnkleRhaukus

(assuming the limit exists of course)

18. anonymous

If $$\space \lim \limits_{x\rightarrow a^-}f(x)=\lim \limits_{x \rightarrow a^+}f(x)$$ then the limit exists.