Question

find the limit L. then find delta>0 such that
lf(x) - Ll < 0.01 whenever 0< lx-cl < delta. ---->
lim (4 -(x/2))
x->4

Answer 1

limit as \(x\to 4\) is
\[4-2=2\] right? so you have
\[|4-\frac{x}{2}-2|<0.01\]

Answer 2

or
\[|2-\frac{x}{2}|<0.01\] meaning
\[|\frac{4-x}{2}|<0.01\]so
\[|4-x|<0.02\] will work

Answer 3

oh and it is clear that \(|4-x|=|x-4|\) right, so \(\delta=0.02\) will work for you here

Answer 4

oh how do u prove that l4-xl=lx-4l ?

Answer 5

in english, \(|x-4|\) is the distance between \(x\) and \(4\)
so is \(|4-x|\)

Answer 6

oh so can i show my work like this.... lx-4l=l -1(x-4) l

Answer 7

you can check it if you like, but there is not much to prove.
if \(a-b>0\) then \(|a-b|=a-b\) and \(|b-a|=-(b-a)=a-b\) as well
you can see what happens if \(a-b<0\)

Answer 8

i would take it as a given, and not bother proving it for this problem
i would assume it

Answer 9

how bout if i'm trying to solve l (-x/2) +2 l < 0.01

Answer 10

can i do this: l (-1/2)(x-4) l < 0.01 and then (1/2) lx-4l < 0.01

Answer 11

? :/

Answer 12

sure

Answer 13

or
\[|2-\frac{x}{2}|=|\frac{4-x}{2}|=\frac{1}{2}|4-x|\] any way you like

Answer 14

oh yeah but that's if you're trying to match the absolute value to l4-xl....?