Evaluate the limit of [square root(x-1)-2]/(x-5) x-->5

In order to solve this limit, we'll check if it is an
indetermination case.

We'll substitute x by 5 in the
expression of the limit:

lim [sqrt(x-1)-2]/(x-5)
= [sqrt(5-1)-2]/(5-5) = (2-2)/(5-5) = 0/0

Since, we've
obtained "0/0", we'll use L'Hospital rule.

This rule claims
that if we deal with an indetermination case, "0/0" or "inf/inf", we can evaluate the
limit in this way:

lim [f(x)/g(x)] = lim {[f(x)]' /
[g(x)]'} if and only if

lim f(x) = 0 and lim g(x) =
0

or

lim f(x) = inf and lim
g(x) = inf

If we substitute x by 7, we'll have an
indetermination case "0/0".

Let's denote f(x) =
sqrt(x-1)-2 and g(x) = x-5

Let's apply L'Hospital rule
now:

lim {[sqrt(x-1)-2]/(x-5)}=lim
{[sqrt(x-1)-2]'/(x-5)'}

[sqrt(x-1)-2]' = (x-1)'/2*sqrt(x-1)
= 1/2*sqrt(x-1)

(x-5)'=1

lim
{[sqrt(x-1)-2]'/(x-5)'}=lim[1/2*sqrt(x-1)]

lim
{[sqrt(x-1)-2]'/(x-5)'}=1/2*sqrt(5-1)

lim
{[sqrt(x-1)-2]/(x-5)} = 1/4