Day 5: Discontinuity Problem

1. Determine all points, if any, at which each of the following functions is discontinuous, and state the type of discontinuity. Justify your answer with an appropriate analysis:
(a)
 
(b)





Solution:
(a) To determine where f(x) is discontinuous, it's necessary to know what could make it discontinuous. Dividing by 0, taking the square root of a negative number are ways a graph can be discontinuous. The only one that applies to f(x) is dividing by zero. Suppose x=2, then 0/0 is obtained:



To identify which type of discontinuity this is, you must know how each type of discontinuity occurs. When dividing by zero, there are two possible discontinuities. If, at the number that causes the denominator to be zero, the numerator is also zero, there is a removable discontinuity called a "hole". If the numerator is nonzero, but the denominator is zero, then the function has an "asymptote". We found that f is discontinuous at x=2 and that f(2)=0/0, so f has a hole at 2.


(b) Now g(x) has a square root and is a fraction, so there are two ways for g to be discontinuous. First of all, let's determine the following limit:



To do so, multiply the top and bottom by the conjugate of the numerator (which is the same as multiplying  by one, so it doesn't change the value of the limit!). The conjugate is simply the numerator, but change the minus to a plus:





Hence, the following is true:



i.e., there is a hole at x=16. 


Finally, if x is negative (x<0), then the square root of x is undefined. Thus, x<0 is not in the domain of g. 


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