We have to find the value of lim x --> 1 [( 1- sqrt x)/ (1- x)]

lim x --> 1 [( 1- sqrt x)/ (1- x)]

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

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

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

substitute x = 1

=> 1/(1 + 1)

=> 1/2

**The required value of the limit is 1/2**

The limit `lim_(x -> 1) ( 1- sqrt x)/ (1- x) ` is required.

Substituting x = 1 in the expression gives the indeterminate form 0/0.

Now `( 1- sqrt x)/ (1- x)` can be written as:

`( 1- sqrt x)/ (1^2- (sqrt x)^2)`

Now use the expansion `a^2 - b^2 = (a-b)(a+b)`

= `( 1- sqrt x)/ ((1+sqrt x)(1- sqrt x))`

= `1/(1+sqrt x)`

Now substitute x = 1 in `1/(1+sqrt x)` , the result is 1/(1+1) = 1/2

The limit `lim_(x -> 1) ( 1- sqrt x)/ (1- x) = 1/2`

We calculate the limit substituting x by 1

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

We notice that we've get an indetermination case

We'll solve using L'Hospital rule:

lim [( 1- sqrt x)/ (1- x)] = lim ( 1- sqrt x)'/ (1- x)'

lim ( 1- sqrt x)'/ (1- x)' = lim (-1/2sqrtx)/-1

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

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

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