Evaluate the limit [sqrt(x+6)-4]/(x-10) x-->10

In order to evaluate this limit, first we'll verify if it
is an indetermination case, and it is, "0/0" case, so, we'll apply L'Hospital
rule.

Let's recall L'Hospital
rule.

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

We'll substitute x by 10 and we'll get an
indetermination case "0/0".

Let's denote f(x) =
sqrt(x+6)-4 and g(x) = x-10

Let's apply L'Hospital rule
now:

lim {[sqrt(x+6)-4]/(x-10)}=lim
{[sqrt(x+6)-4]'/(x-10)'}

[sqrt(x+6)-4]' =
(x+6)'/2*sqrt(x+6) =
1/2*sqrt(x+6)

(x-10)'=1

lim
{[sqrt(x+6)-4]'/(x-10)'}=
lim[1/2*sqrt(x+6)]=1/2*sqrt(10+6)

lim
{[sqrt(x+6)-4]/(x-10)} =
1/2sqrt16

lim
{[sqrt(x+6)-4]/(x-10)} = 1/8