Question

Let f(x) = { 4-x^2 if x is less tahn or = to 2 ; x-1 if x is greater than 2}

a) find lim as x appraches 2 from the left f(x) and lim as x appraches 2 from teh right f(x)

Answer 1

\[a) \lim_{x \rightarrow 2-} f(x) and \lim_{x \rightarrow 2+} f(x)\]

Answer 2

a) As you approach it from the left, you put in values x<2, so stick with the first branch, so you get 4-2^2=0. As you approach from the right, you put in values x>2, so you get 6-2=4.

Answer 3

b) does \[\lim_{x \rightarrow 2} f(x) \] exist ?

c) what would te hgraph of f look like?

Answer 4

b) No

Answer 5

c) It makes a jump from 0 to 4 at x=2.

Answer 6

how come b does not exist

Answer 7

because left side and right side limits don't match up

Answer 8

oh yeah i see 0 and 4 dont match

Answer 9

lim f(x) x -> 2_
= lim 4 - x^2
= lim 4 - lim x^2
= lim 4 - (lim x)^2
= 4 - 2^2
= 4 -4
= 0
lim f(x) x-> 2+
= lim (x-1)
= lim x - lim(1)
= 2 - 1
= 1

lim 2_ not equal to lim 2+ since 0 <> 1, therefore lim x x -> 2 does not exist.

Answer 10

thanks alot... =)

Answer 11

np