How do you use the formal definition of a limit to prove #lim (x/(x-3)) =1# as x approaches infinity?

Question

Samantha Adkison · Accepted Answer

# lim_(x->oo)x/(x-3) = 1#

If we look at the graph of #y=x/(x-3)# we can see that it is clear that the limit exists, and is approximately #1#

graph{x/(x-3) [-30, 30, -2, 2]}

Now, As #x->oo# then #1/x->0#

So, it would be better if we could replace #x# with #1/x#

# lim_(x->oo)x/(x-3) = lim_(x->oo)x/(x-3)*(1/x)/(1/x) #

# :. lim_(x->oo)x/(x-3) = lim_(x->oo)(1/x x)/(1/x(x-3)) #

# :. lim_(x->oo)x/(x-3) = lim_(x->oo)(1)/(1-3/x) #

And, using #1/x->0# as #x->oo# we have;

# :. lim_(x->oo)x/(x-3) = (1)/(1-0) = 1# QED

Which is completely consistent with the above graph.

Emily Ammerman · Answer

Please see below.

I take the formal defintion to be: #lim_(xrarroo)f(x) = L# if and only if for every positive #epsilon#, there is an #M# such that for all #x#, if #x > M#, then #abs(f(x)-L) < epsilon# Preliminary investigation To show that #lim_(xrarroo)x/(x-3) =1#, we need to make #abs(x/(x-3) - 1)# small, by making #x# large. #abs(x/(x-3) - 1) = abs (x/(x-3) - (x-3)/(x-3)) = abs(3/(x-3)) = 3/abs(x-3)# If we make sure that #M > 3#, the we will have # abs(x/(x-3) - 1) = 3/(x-3)# So we need to make #3/(x-3) < epsilon# and we will make sure that #x-3# is positive, so we'll be OK if we make #3/epsilon < x-3# and #3/epsilon+3 < x#. So our #M# will be #3/epsilon+3# (or a number greater than that). Claim: #lim_(xrarroo)x/(x-3) =1# Proof: Given #epsilon > 0#, let #M = 3/epsilon+3# (Note that #M > 3# since #3/epsilon# is positive.) For any #x > M#, we have #x -3 > 3/epsilon#, so #x-3 > 0# and #epsilon (x-3) > 3# and, finally #epsilon > 3/(x-3)# Observe, now, that for #x > M#, #abs(3/(x-3)-1) = abs(3/(x-3)) = 3/(x-3) < epsilon# Therefore, by the definition of limit at infinity, #lim_(xrarroo)x/(x-3) =1#.

Emma Aldridge · Answer

undefined To use the formal definition of a limit to prove that lim (x/(x-3)) = 1 as x approaches infinity, we need to show that for any positive number ε, there exists a corresponding positive number M such that if x is greater than M, then |(x/(x-3)) - 1| is less than ε.

Let's start by simplifying the expression (x/(x-3)):
(x/(x-3)) = (x/(x(1-3/x))) = 1/(1-3/x)

Now, we want to find an M such that if x is greater than M, then |(1/(1-3/x)) - 1| is less than ε.

To simplify further, we can multiply the numerator and denominator of the expression by x:
1/(1-3/x) = x/(x-3x)

Now, we can rewrite the expression as:
1/(1-3/x) = x/(x-3x) = x/(-2x) = -1/2

Since -1/2 is a constant, it does not depend on x. Therefore, it is clear that as x approaches infinity, the expression (x/(x-3)) approaches -1/2, not 1.

Hence, using the formal definition of a limit, we cannot prove that lim (x/(x-3)) = 1 as x approaches infinity.