I am not sure exactly what is meant by "algebraically" in the question. So I will try to show this by a couple of different approaches.

**Method 1**

The simplest approach would be to use the properties of continuous functions. A standard result is that the ratio #f(x)(/g(x)# of two continuous functions #f(x)# and #g(x)# is continuous at #x=a#, provided #g(a) ne 0#.

Since #x+3# and #x-3# are obviously continuous functions (you may prove this statement, if proof is needed, using the property that the sum of continuous functions is continuous), and #x-3# does not vanish at #x=4#, the ratio #(x+3)/(x-3)# is continuous at #x=4#.

Hence, the limit as #x to 4# of #(x+3)/(x-3)# is equal to the value of #(x+3)/(x-3)# at #x=4#, namely #color(red)((4+3)/(4-3) = 7)#

**Method 2**

Another approach could be to use the #epsilon-delta# definition of the limit. To show that the limit is 7 according to this definition we have to show that

#forall epsilon >0, exists delta > 0 : |x-4| < delta implies |(x+3)/(x-3)-7|< epsilon #

Now

#|(x+3)/(x-3)-7| = |(-6x+24)/(x-3)| = 6|x-4|/|x-3|#

For #|x-4|< delta < 1/2# we have #|x-3|>1/2# and so

#|(x+3)/(x-3)-7| < 6 delta /(1/2)=12 delta#

So, for any #epsilon>0# we will have

#|(x+3)/(x-3)-7| < epsilon#

if we choose

#|x-4| < delta = min{epsilon/12,1/2}#