The value of lim x-->3 [sqrt ( x^2 - 9)/(x - 3)] has to be found.

lim x-->3 [sqrt ( x^2 - 9)/(x - 3)]

substituting x = 3, gives the indeterminate form 0/0.

=> lim x-->3 [sqrt (x - 3)* sqrt (x + 3)/(x - 3)]

=> lim x-->3 [sqrt (x + 3)/ sqrt (x - 3)]

substituting x = 3 now gives sqrt 6 / 0 which is infinity.

**The required limit is infinity.**

The limit `lim_(x=->3) sqrt( x^2 - 9)/(x - 3)` has to be determined. If x is substituted with 3, the result is of the form 0/0 which is not defined.

l'Hospital's rule can be used to find the limit. For this replace the numerator and denominator by their derivatives.

This gives `lim_(x->3) ((1/2)*2x*(1/sqrt(x^2 - 9)))/1`

= `lim_(x->3) (x*(1/sqrt(x^2 - 9)))/1`

Now substitute x = 3, we get 3/0, this is equal to infinity as the result of dividing any number by 0 gives infinity.