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

2 Answers
Nov 3, 2016

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

Explanation:

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.

Nov 13, 2016

Please see below.

Explanation:

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#.