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

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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...

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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`

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