How do you prove that the limit of #x^2 = 9# as x approaches 3 using the epsilon delta proof?

Question

Cameron Alexander · Accepted Answer

See below.

Prove #lim_(x->3)x^2=9#. The epsilon delta proof for limits is easier understood when one is familiar with the definitions of the terms involved. Most useful will be the definition of the limit of a function. #lim_(x->c)f(x)=L# Definition: for every #epsilon>0#, there exists #delta>0# such that #abs(f(x)-L)< epsilon# whenever #0< abs(x-c)< delta#. In our case: We then need to find an expression for #delta# so that the definition, and consequently the proof, holds. In other words, we need: #abs(x^2-9)< epsilon" "# when #" "0< abs(x-3)< delta# So: #abs(x^2-9)< epsilon# #=>abs((x-3)(x+3))< epsilon# Since the concept of the limit only applies when #x# is close to #a#, we will need to restrict #x# so that it is at most #1# away from #a#, or: #abs(x-a) <1#. For us, that is #abs(x-3)< 1#. #=>-1< (x-3) <1# #=>2< x <4# #=>5< (x+3) <7# #=> (x+3) <7# We are looking at the maximum values for both cases. So we know that we must have: #abs((x-3)(x+3))< 7delta=epsilon# #=>delta=epsilon/7# Then our two restrictions are: #abs(x-3)<1# and #abs(x-3)< delta/7# We want a #delta# which will satisfy both inequalities, so we'll end up taking the minimum value for #epsilon# between #1# and #epsilon/7#. Note that all of the above steps come from the definition of the limit of a function as provided above. Proof. Let #epsilon>0# be arbitrary and let #delta=min{1,(epsilon)/7}#. If #abs(x-3)< delta#, then #abs(x-3) <1#; hence: #-1< x-3 <1# #=>2< x< 4# #=>5< x+ 3< 7# #=>abs( x+3)< 7# Also, if #abs(x-3)< delta#, then #abs(x-3)< epsilon/7#, so: #abs(x^2-9)=abs((x-3)(x+3))# #<7*abs(x-3)# #<=7*epsilon/7=epsilon" " square# Another way to go about working out the scratch work is as follows: We know we need #abs(x^2-9)< epsilon# #=>abs((x-3)(x+3))< epsilon# #=>abs(x-3)< epsilon/(abs(x+3))# #=>delta < epsilon/(abs(x+3))# #abs(x-3)<1# #=>-1< abs(x-3) < 1# #=>2< x < 4# #=> 5< (x+3) <7# So, for #abs(x-3)< epsilon/(abs(x+3))# we can see that the RHS is at a minimum when #(x+3)# is at a maximum, i.e. the LHS decreases as the RHS increases. The maximum of #x+3# was found above to be #7#. #:. delta=epsilon/7#

Ezekiel Alexander · Answer

undefined To prove that the limit of x^2 is 9 as x approaches 3 using the epsilon-delta proof, we need to show that for any given positive value of epsilon, there exists a positive value of delta such that if the distance between x and 3 (denoted as |x - 3|) is less than delta, then the distance between x^2 and 9 (denoted as |x^2 - 9|) is less than epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a delta > 0 such that if 0 < |x - 3| < delta, then |x^2 - 9| < epsilon. We start by assuming 0 < |x - 3| < delta. Expanding the expression |x - 3|, we have -delta < x - 3 < delta. Adding 3 to all parts of the inequality, we get -delta + 3 < x < delta + 3. Now, let's consider the expression |x^2 - 9|. Expanding it, we have |x^2 - 9| = |(x - 3)(x + 3)|. Since we have assumed 0 < |x - 3| < delta, we can also assume that 0 < x - 3 < delta, and thus 0 < x < delta + 3. Now, we can rewrite |x^2 - 9| as |(x - 3)(x + 3)| < (delta + 3)(x + 3). To ensure that |x^2 - 9| < epsilon, we need to choose a suitable value for delta. Let's choose delta = min(1, epsilon/6). Now, if 0 < |x - 3| < delta, we have 0 < x < delta + 3, which implies 0 < x < 1 + 3 = 4. Using this information, we can rewrite (delta + 3)(x + 3) as (delta + 3)(4 + 3) = 7(delta + 3). Since we chose delta = min(1, epsilon/6), we have delta ≤ 1, which implies 7(delta + 3) ≤ 7(1 + 3) = 28. Therefore, if 0 < |x - 3| < delta, we have |x^2 - 9| < 28. To satisfy the condition |x^2 - 9| < epsilon, we need to ensure that 28 < epsilon. Hence, by choosing delta = min(1, epsilon/6), we have shown that if 0 < |x - 3| < delta, then |x^2 - 9| < epsilon, proving that the limit of x^2 is 9 as x approaches 3 using the epsilon-delta proof.