Question

How do you find the limit of `(x+5)/(25-x^2)` as x approaches `5^-`?

Answer 1

It is

#lim_(x->5^-) (x+5)/(25-x^2)=lim_(x->5^-) (x+5)/((5-x)*(5+x))= lim_(x->5^-) cancel(x+5)/((5-x)*cancel(5+x))= lim_(x->5^-) 1/(x-5)=-oo#