# Evaluate the limit : limit [ (sqrtx - 1 )/(x^2 - 1 ) ] . x-> 1

hala718 | Certified Educator

lim[(sqrtx-1)/(x^2-1)]    x--> 1

When substituting with x=1 , the limit is 0/0.

That means that (1) is a root for the numerator and the denominator that x=1 is a common factor. So, we need to to factorize in order to reduce the common factor:

lim(sqrtx-1)/(x-1)(x+1) = lim(sqrtx-1)/(sqrtx-1)(sqrtx+1)(x+1)

Reduce similar:

==> lim 1/(sqrtx+1)(x+1) , x--> 1

==> lim 1/(1+1)(1+1) = 1/4

neela | Student

To find the limit,  [ (sqrtx - 1 )/(x^2 - 1 ) ] . x-> 1

Solution:

Lt (sqrt x-1)/(x^2-1) as x-->1, is like a 0/0 indeterminate form.  But  the denominator x^2-1 = (x+1)(x-1) = (x+1)[x^(1/2)-1][x^(1/2)+1].

So the given lt is :

Lt [x^(1/2) - 1]/{ (x+1) (x^(1/2)-1) (x^(1/2)+1)} as x-->1

= Lt 1/[(x+1) (x^(1/2)+1)], as x-->1

= 1/[(1+1)((1^(/2)+1)]

= 1/4

giorgiana1976 | Student

We have to calculate lim [(sqrt x-1)/(x^2-1)] and then we'll do the steps:

-we'll write the difference of squares as:

x^2 - 1 = (x-1)(x+1), where x-1 could be considered as a difference of squares also, x-1 = (sqrtx-1)(sqrtx+1)

- after reducing similar terms, we can calculate the limit:

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

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

We'll just simply substitute x by 1, into the limit:

lim 1/(sqrtx+1)(x+1) = 1/(1+1)(1+1)=1/2*2=1/4

tonys538 | Student

The limit limit `lim_(x->1)(sqrtx - 1 )/(x^2 - 1 )` has to be determined.

If we substitute x = 1 the result `(sqrt 1 - 1 )/(1^2 - 1 ) = 0/0` . This is an indeterminate form and allows the use of l'Hospita'ls rule. Substitute the numerator and denominator with their derivative.

The limit is now changed to:

`lim_(x->1) ((1/2)*(1/sqrt x))/(2x)`

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

= `lim_(x->1) 1/(4*x*sqrt x)`

Again substituting x = 1 gives the result 1/4

The required value of `lim_(x->1)(sqrtx - 1 )/(x^2 - 1 ) = 1/4`