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}
