Question

lim (4- (x/2))
x->4

I know the limit is L=2 but how do I solve further!

Answer 1

since the function is continuous at x=4 we can just plug in

Answer 2

Right, but how do i get to \[\delta=\epsilon / ?\]

Answer 3

let me see if i can write this out

Answer 4

ok we can do this. do you know what the goal is?

Answer 5

all I know is that im supposed to find an answer in that form :(

Answer 6

you say the limit is 2, so that means that you can force
\[4-\frac{x}{2}\] arbitrarily close to 2 if you make x close to 4. in math we say given any epsilon we can make
\[|4-\frac{x}{2}-2|<\epsilon\] by making
\[|x-4|<\delta\] and of course we write
\[\delta\] in terms of
\[\epsilon\]

Answer 7

so you work backwards. you start with
\[|4-\frac{x}{4}-2|<\epsilon\] and see what you need. mostly it is algebra

Answer 8

\[|4-\frac{x}{2}-2|=|2-\frac{x}{2}|=|\frac{4-x}{2}|=\frac{1}{2}|4-x|<\epsilon\]

Answer 9

this is good because we have control over
\[|x-4|=|4-x|\] we just have to make sure that it is small enough so that the above inequality holds
we can pick
\[\delta = 2\epsilon\] and then we do the work backwards again.

Answer 10

if
\[|4-x|<2\epsilon\] then
\[|4-\frac{x}{2}-2|=|2-\frac{x}{2}|=|\frac{4-x}{2}|=\frac{1}{2}|4-x|<\epsilon\] as needed

Answer 11

So would the final answer be epilon / ?

Answer 12

i sense you might be confused about this. the "final answer" is for you to find delta in terms of epsilon, that is, someone says "make it within epsilon of 2" and you say, ok, make
\[|x-4|<2\epsilon\] it is your job to find the delta that works, and in this case
\[\delta = 2\epsilon\] is all that is needed

Answer 13

if you are claiming the limit as x approaches a of f(x) is L (2 in this case) you have to show that given ANY epsilon you can come up with a delta so that if
\[|x-a|<\delta\] then
\[|f(x)-L|<\epsilon\]

Answer 14

so basically you are done when you find delta in terms of epsilon