How do you prove the statement lim as x approaches 9 for #root4(9-x) = 0# using the epsilon and delta definition?

The #delta"-"epsilon# definition of a limit states that:

the limit of a function #f(x)#, as #x# approaches some value #c#, is #L# if, for every possible #epsilon>0#, we can find a #delta>0# that depends on #epsilon#, such that #abs(f(x)-L) < epsilon# whenever #abs(x-c) < delta#.

It's like a game. Player 1 picks an #epsilon>0#, and Player 2 is trying to find a #delta>0# such that every #x# within #+-delta# of #c# (that is, #color(navy)(x in (c-delta, c+delta))#) is guaranteed to get mapped to an #f(x)# within #+-epsilon# of #L# (that is, #color(green)(f(x) in (L-epsilon, L+epsilon))#). Player 1 keeps picking smaller and smaller #epsilon#, and Player 2 keeps having to find smaller and smaller #delta#.

If we can prove that, for every #epsilon# Player 1 picks, Player 2 can find a suitable #delta#, then we've proven the limit. In other words:

If

#AA" "epsilon > 0" "EE" "delta>0"#

such that

#abs(f(x)-L) < epsilon # when #abs(x-c) < delta#

then

#lim_(x->c)f(x)=L#.

Okay, so that's a lot of mumble jumble. Let's put it to use.

We need to find a #delta# that depends on #epsilon#. That means you can think of #delta# as a function of #epsilon#. We also want to have #'x " is within "delta" of "c'# imply #'f(x)" is within " epsilon " of "L'#, or in math lingo:

#abs(x-c) < delta => abs(f(x)-L) < epsilon#

So we start with #abs(x-c) < delta# and #color(blue)("try to get it to look like " abs(f(x)-L) < epsilon)#.

In this example, #c=9#, #L=0#, and #f(x)=root(4)(9-x)#.

#abs(x-c) < delta => abs(x-9) < delta#
#color(white)(abs(x-c) < delta) => abs(9-x) < delta#
#color(white)(abs(x-c) < delta) => root(4)abs(9-x) < root(4)delta#
#color(white)(abs(x-c) < delta) => abs(root(4)(9-x)) < root(4)delta#
#color(white)(abs(x-c) < delta) => abs(root(4)(9-x)-0) < root(4)delta#
#color(white)(abs(x-c) < delta) => abs(f(x)-0) < root(4)delta#

Hey—looks like we may have found our connection between #delta# and #epsilon#! If we let #root(4)delta = epsilon#, then we have

#color(white)(abs(x-c) < delta) => abs(f(x)-0) < epsilon#

and so we've shown that #abs(x-9) < delta => abs(f(x)-0) < epsilon#.

The last thing to do is to solve #root(4)delta = epsilon# for #delta#:

#"   "root(4)delta=epsilon#
# => delta = epsilon^4#

All the work we've done here simply means that no matter how small an #epsilon# Player 1 may pick, Player 2 can always just choose their #delta# to be #epsilon^4#, and they'll win every time. That is, as long as #x# is within # epsilon^4# of #9#, #root(4)(9-x)# will be within #epsilon# of #0#. Thus,

#lim_(x->9)root(4)(9-x)=0#

has been proven.

#QED.#