How do you find the limit #(sqrtx-1)/(x-1)# as #x->1#?
3 Answers
The limit is
Explanation:
We use l'Hôpital's Rule
Limit
So this is impossible, so we apply l'Hôpital's Rule
Limit
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#
Explanation:
so
