lim_(x->5) [sqrt(x-1)-2]/(x-5) = lim_(x->5) (x-1-4)/((x-5)(sqrt(x-1)+2) (multiplication by the conjugate (sqrt(x-1)+2) )

lim_(x->5) [sqrt(x-1)-2]/(x-5) = lim_(x->5) (x-5)/((x-5)(sqrt(x-1)+2)

lim_(x->5) [sqrt(x-1)-2]/(x-5) = lim_(x->5) 1/((sqrt(x-1)+2)

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

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

lim_(x->5) 1/((sqrt(x-1)+2) = 1/4

**Hence, evaluating the limit to the function yields lim_(x->5) [sqrt(x-1)-2]/(x-5) = 1/4.**

We need to find lim x--> 5 [(sqrt(x - 1) - 2)/(x-5)]

substituting x = 5 we get the form 0/0 which is indefinite, so we can use L'Hopital's rule and substitute the numerator and denominator with their derivatives

=> lim x--> 5[ (1/2)(x - 1)^(-1/2)]

substitute x = 5

=> (1/2)(1/2)

=> 1/4

**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