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Hi

I did post this question before but removed it as i typed it incorrectly when i posted it .

i have posted a similar question but this one is different.
I would like to find out how to solve this problem,

Let f(x) =6x+4 if x is less than 3 and also equal sqrt 6(x-3) if x is greater than or equal to 3

then why is the function not continuous out of the following below

though both left and right limits exist at x = 3, they are not equal.

the right limit does not exist at x = 3.

though the limit at x = 3 exists, the value of the limit is not equal to the value of the function at x = 3

the left limit does not exist at x = 3.

None of the above

thanks for your help, it is much appreciated :)

2007-04-10 19:10:03 · 2 answers · asked by zz06 3 in Science & Mathematics Mathematics

2 answers

As x -> 3 from below, f(x) -> 6(3) + 4 = 22.
As x -> 3 from above, f(x) -> sqrt 6(0) = 0. (Note that sqrt 6(x-3) is continuous on [3, ∞) so we can just take the value of the function at 3 here.)
So the left and right limits exist at x=3, but are not equal to each other.

2007-04-10 19:14:31 · answer #1 · answered by Scarlet Manuka 7 · 0 0

though both left and right limits exist at x = 3, they are not equal:

left limit = sqrt(6*(3-3))=0,
right limit = 6*3+4=22

2007-04-10 19:28:08 · answer #2 · answered by dd4dd2dd1 2 · 0 0

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