We have to find [square root(x-1)-2]/(x-5) lim x-->5.

Here we cannot replace x with 5 as we get a result of the form 0/0 which cannot be determined.

Therefore do the following.

[sqrt(x-1) -2] / (x-5)

=> [sqrt(x-1) -2] / (x-1 -4 )

Now we take (x-1 -4 ) as (sqrt (x-1))^2 - 2^2 which is equal to [(sqrt (x-1) - 2)(sqrt (x-1) + 2)]

=> [sqrt(x-1) -2] / [(sqrt (x-1) - 2)(sqrt (x-1) + 2)]

cancelling [sqrt(x-1) -2]

=> 1/ (sqrt (x-1) + 2)

now equate x to 5

=> 1/ (sqrt (5-1) + 2)

=> 1/ (sqrt (4) + 2)

=> 1/ (2 +2)

=> 1/4

**Therefore the required limit is 1/4**

In order to solve this limit, we'll check if it is an indetermination case.

We'll substitute x by 5 in the expression of the limit:

lim [sqrt(x-1)-2]/(x-5) = [sqrt(5-1)-2]/(5-5) = (2-2)/(5-5) = 0/0

Since, we've obtained "0/0", we'll use L'Hospital rule.

This rule claims that if we deal with an indetermination case, "0/0" or "inf/inf", we can evaluate the limit in this way:

lim [f(x)/g(x)] = lim {[f(x)]' / [g(x)]'} if and only if

lim f(x) = 0 and lim g(x) = 0

or

lim f(x) = inf and lim g(x) = inf

If we substitute x by 7, we'll have an indetermination case "0/0".

Let's denote f(x) = sqrt(x-1)-2 and g(x) = x-5

Let's apply L'Hospital rule now:

lim {[sqrt(x-1)-2]/(x-5)}=lim {[sqrt(x-1)-2]'/(x-5)'}

[sqrt(x-1)-2]' = (x-1)'/2*sqrt(x-1) = 1/2*sqrt(x-1)

(x-5)'=1

lim {[sqrt(x-1)-2]'/(x-5)'}=lim[1/2*sqrt(x-1)]

lim {[sqrt(x-1)-2]'/(x-5)'}=1/2*sqrt(5-1)

**lim {[sqrt(x-1)-2]/(x-5)} = 1/4**