lim_(x->5^(+)) sqrt(25-x^2)/(x-5) Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve,
lim_(x->5^(+)) sqrt(25-x^2)/(x-5)
 
Removing the negative form the denominator we get
 
= - lim_(x->5^(+)) sqrt(25-x^2)/(5-x)
= - lim_(x->5^(+)) sqrt(5^2-x^2)/(5-x)
= - lim_(x->5^(+)) sqrt((5-x)(5+x))/(5-x)
= - lim_(x->5^(+)) sqrt((5+x)/(5-x))
=- [(lim_(x->5^(+)) sqrt((5+x))]/ [lim_(x->5^(+)) sqrt(5-x))]
= - sqrt(5+lim_(x->5^(+)) x) /sqrt(5-lim_(x->5^(+)) x)
as x->  5^(+) ,then the denominator tends from 0 to -1 .
so,
lim_(x->5^(+)) sqrt(25-x^2)/(x-5)=oo