How do you prove that the limit # 3/sqrt(x-5) = ∞# as x approaches #5^+# using the formal definition of a limit?

1 Answer
May 7, 2017

Please see below.

Explanation:

Finding the proof
This explanation of finding the proof is a bit long. If you just want to read the proof, scroll down.

By definition,

#lim_(xrarra^+)f(x) = oo# if and only if

for every #M > 0#, there is a #delta > 0# such that:

for all #x#, if #0 < x-a < delta#, then #f(x) > M#.

So we want to make #f(x)# greater than some given #M# and we control (through our control of #delta#) the size of #x-5#

We want: #3/sqrt(x-5) > M#

Each factor in this inequality is positive, so we can multiply by #sqrt(x-5)# and divide by #M# without changing the direction of the inequality.

We want #3/M > sqrt(x-5)#

In order to make this true, it suffices to make #x-5# positive and less than #9/M^2#.

Note that:
The square root function is an increasing function, that is: if #0 < a < b#, then #sqrta < sqrtb#

Writing the proof

Claim: #lim_(xrarr5^+)3/sqrt(x-5) = oo#

Proof:

Given #M > 0#, choose #delta = 9/M^2#. (Note that #delta# is positive.)

Now if #0 < x-5 < delta# then

#0 < x-5 < 9/M^2#

So #sqrt(x-5) < 3/M#

Therefore, #M < 3/sqrt(x-5)#

We have shown that for any positive M, there is a positive #delta# such that for all #x#, if #0 < x-5 < delta#, then #3/sqrt(x-5) > M#.

So, by the definition of limit from the right and infinite limit, we have #lim_(xrarr5^+)3/sqrt(x-5) = oo#.