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