Question

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

Answer 1

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`.