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

Question

Charles Arocha · Accepted Answer

See explanation.

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#. That's a lot of jumbled text, so let's make use of it. 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 demonstrated. #QED.#

William Abdalla · Answer

undefined To prove the statement lim as x approaches 9 for root4(9-x) = 0 using the epsilon and delta definition, we need to show that for any given epsilon greater than 0, there exists a corresponding delta greater than 0 such that if 0 < |x - 9| < delta, then |root4(9-x) - 0| < epsilon. Let's start by considering |root4(9-x) - 0|. Since root4(9-x) is always non-negative, we can simplify this to root4(9-x). Now, we want to find a delta such that if 0 < |x - 9| < delta, then |root4(9-x) - 0| < epsilon. To do this, we can manipulate the expression |root4(9-x) - 0| < epsilon. We can rewrite |root4(9-x) - 0| < epsilon as root4(9-x) < epsilon. To proceed, we raise both sides of the inequality to the fourth power, giving us (root4(9-x))^4 < epsilon^4. Simplifying further, we have 9-x < epsilon^4. Now, we can choose delta = epsilon^4. If 0 < |x - 9| < delta, then 9-x < epsilon^4, which implies root4(9-x) < epsilon. Therefore, we have shown that for any given epsilon > 0, there exists a corresponding delta > 0 (specifically, delta = epsilon^4) such that if 0 < |x - 9| < delta, then |root4(9-x) - 0| < epsilon. This proves the statement lim as x approaches 9 for root4(9-x) = 0 using the epsilon and delta definition.