Verified answer

limit x->3 √(x+3) = √6

If the function is: [involving simple algebraic cancellation]

√((x^2-9)/(x-3)) and not [√(x^2 - 9)]/(x-3);

then it is the following:

lim x --> 3 √((x+3)(x-3)/(x-3))

= lim x--> 3 √(x+3)

= √(3+3) = √6

If the function is [√(x^2-9)]/(x-3), then we use l'hoptial's rule:

lim (x->c) f(x)/g(x) = lim (x->c) f '(x)/g '(x)

This can only be used if, when you plug in the limit value of c, an indeterminate form is produced; an indeterminate form is 0/0, or infinity/infinity, or infinity/0...

lim(x->3-) ((1/2)*(x^2-9)^(-1/2)*2x)/(1)

= lim(x->3-) (x*(x^2-9)^(-1/2))

= 3*0^(-1/2) = 3/0

= Infinity

But, the value is probably from the first limit; so, the value is √6 of the limit. If you do not know l'hopital's rule, then you should definitely learn to use it.

There you go. Hope that helps.