How do you find the limit #(sqrtx-1)/(x-1)# as #x->1#?

3 Answers
Narad T.
Oct 29, 2016

The limit is #1/2#

Explanation:

We use l'Hôpital's Rule

Limit #(sqrtx-1)/(x-1)=0/0#
#x->1#
So this is impossible, so we apply l'Hôpital's Rule

Limit #(sqrtx-1)/(x-1)=((sqrtx-1)')/((x-1)')=1/(2sqrtx)*1/1=1/2#
#x->1#

George C.
Oct 29, 2016

#lim_(x->1) (sqrt(x)-1)/(x-1) = 1/2#

Explanation:

Note that:

#(sqrt(x)-1)(sqrt(x)+1) = x-1#

Hence:

#lim_(x->1) (sqrt(x)-1)/(x-1) = lim_(x->1) color(red)(cancel(color(black)(sqrt(x)-1)))/(color(red)(cancel(color(black)((sqrt(x)-1))))(sqrt(x)+1))#

#color(white)(lim_(x->1) (sqrt(x)-1)/(x-1)) = lim_(x->1) 1/(sqrt(x)+1)#

#color(white)(lim_(x->1) (sqrt(x)-1)/(x-1)) = lim_(x->1) 1/(1+1)#

#color(white)(lim_(x->1) (sqrt(x)-1)/(x-1)) = 1/2#

Cesareo R.
Oct 29, 2016

#1/2#

Explanation:

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

so

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

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
2042 views around the world
You can reuse this answer
Creative Commons License