Question

Find the limit as x approaches 2

for (4-x/2), then find delta > 0 such that |f(x)-L| < 0.01 whenever 0 < |x - c | < delta

I could use 1) step by step explanation
2) a resource other than khan academy for delta epsilon examples

Answer 1

since there is not bad values for x, it can be anything it wants to be and have a good value, then the limit is the same as the value of the equation when x = 2; which is either 3 or 1 in this case depending on how you meant to write it.

\[\frac{4-(2)}{2}=1\]
\[4-\frac{(2)}{2}=3\]

To determine delta, solve for f(x) = 2+.01 , and f(x) = 2-.01 and pick the "x" value that is smaller

Answer 2

one way is the one given by lifesaver which uses the knowledge that 4-x/2 is a continuous function of x.
if you really want to verify the limit by epsilon delta definition
choose an \[\epsilon\].
now create the inequation
\[\left| 4-x/2-3 \right|<\epsilon\]
solving it you will get \[x >2(1-\epsilon)\] and \[x<2(1+\epsilon)\]

hence our \[\delta=\min(2(1+\epsilon),2(1-\epsilon))=2(1-\epsilon)\]
as for this delta the definition of limit is satisfied .
hence for any \[\epsilon\] we are able to find a \[\delta\] satisfying the definition of limit.
hence the limit exists and is 3.