Question

Prove the statement using the epsilon and delta definition of a limit.

lim as x -> 3 (x^2 +x -4) = 8

Answer 1

we can do this. ready?

Answer 2

first of all like in all these problems we work backwards

Answer 3

okay =)

Answer 4

we want a delta, but first we see how to force
\[x^2+x-4\] to be close to 8, but which we mean
\[|x^2+x-4-8|<\epsilon\]

Answer 5

so we are really looking at
\[|x^2+x-12|<\epsilon\] and we want a delta that will make this true if
\[|x-3|<\delta\]

Answer 6

first we factor and see
\[|x^2+x-12|<\epsilon\] is the same thing as
\[|(x-3)(x+4)|<\epsilon\] and now this is not too bad because we have control over the
\[(x-3)\] factor. unfortunately we also have
\[(x+4)\] to contend with.

Answer 7

so the gimmick here is to say that lets make sure that
\[|x-3|<1\] say. this will give a bound on
\[|x+4|\] don't forget we get to control
\[|x-3|\] so we can make it as small as we like and we can certainly make it smaller than 1

Answer 8

now if
\[|x-3|<1\] that forces
\[-1<x-3<1\] or
\[2<x<4\] which means therefore
\[|x+4|<8\]

Answer 9

so we say at the outset that
\[|x-3|<1\] which means
\[|(x-3)(x+4)|<8|x-3|\] and we want
\[|(x-3)(x+4)|<\epsilon\]
\[\delta =\frac{\epsilon}{8}\]
and therefore if
\[|x-3|<\frac{\epsilon}{8}\]we have making
\[|(x-3)(x+4)|<8|(x-3)|<8\times \frac{\epsilon}{8}=\epsilon\]

Answer 10

wow!! taht was a lot of work!!! is this the end of the problem?

Answer 11

that is it. all steps are there for you to look at. hope it becomes clear more or less

Answer 12

YES..THANK U!!! im hopeless when it comes to limits...i do not understand them very well. can u help me with a proof of a limit??

Answer 13

another one?

Answer 14

post and i will see what we can do

Answer 15

okay