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

#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#

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

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#.

To prove that the limit of 3/sqrt(x-5) as x approaches 5 from the positive side is infinity using the formal definition of a limit, we need to show that for any positive value M, there exists a corresponding positive value δ such that if 0 < x - 5 < δ, then 3/sqrt(x-5) > M. Let's proceed with the proof: Given M > 0, we want to find a δ > 0 such that if 0 < x - 5 < δ, then 3/sqrt(x-5) > M. To begin, let's consider the expression 3/sqrt(x-5). We can rewrite it as 3 * (1/sqrt(x-5)). Now, we want to find a δ such that if 0 < x - 5 < δ, then 3 * (1/sqrt(x-5)) > M. To simplify the expression further, we can multiply both the numerator and denominator of 1/sqrt(x-5) by sqrt(x-5), which gives us 3 * (sqrt(x-5)/(x-5)). Now, we need to find a δ such that if 0 < x - 5 < δ, then 3 * (sqrt(x-5)/(x-5)) > M. Since we are approaching 5 from the positive side, we can assume that x > 5. Therefore, x - 5 > 0. Now, let's consider the expression 3 * (sqrt(x-5)/(x-5)). We can see that as x approaches 5 from the positive side, the denominator (x-5) approaches 0, while the numerator (sqrt(x-5)) remains positive. Since the numerator remains positive and the denominator approaches 0, the entire expression 3 * (sqrt(x-5)/(x-5)) will approach positive infinity as x approaches 5 from the positive side. Therefore, we have proven that the limit of 3/sqrt(x-5) as x approaches 5 from the positive side is indeed infinity using the formal definition of a limit.