For the function, determine all values at which the function is discontinuous.
The denominator factors, therefore we could rewrite this as
This function is undefined, and therefore discontinuous, at x = 4 and x = 5.The only place limits could go wrong, besides x = 4 and x = 5, is when x = 1. At x = 1 we have
As x approaches 1 from the left we have
As x approaches 1 from the right we have
Since the one-sided limits agree,
This agrees with h(1) = 1,
so the function is continuous at x = 1.
To summarize, h is only discontinuous at x = 4 and x = 5.