Question

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

Answer 1

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`

Answer 2

`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`

Answer 3

`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`