# `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...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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`

Approved by eNotes Editorial Team