Applied Calculus

Question

Give an example of a non-piecewise function (so it should only have one formula over its entire

domain) that is not continuous at x = 0.

Answer 1

Hi Heena - here’s a hint and a solution. Please note that 1/2 of Paul’s answer is incorrect. The function sin(x) / x is continuous at x = 0.

Hint: one way to cook up such a function is to select one that evaluates to a non-zero number / zero, when you plug in 0. Such a function is said to be unbounded at the point 0. This is why Paul’s example 1/x is in fact, discontinuous at 0. But it also partially illustrates why his other example, sin(x)/x, is continuous at x=0. I say partial because the full reason requires examination of something called the order of the singularities of the function which is beyond the scope of this problem.

Solution: take any function that’s non-zero at x=0. Some examples are cos(x), e^x, 1/x^2, and |x +1|. Now divide each of them by x, e.g. cos(x)/x, etc. All of these functions are now discontinuous at x=0.

Another approach is to identify a function that is unbounded at some finite point. For instance, log(x) is unbounded and technically not defined at x=0 when x is a real number. So log(x) is another example. The function tan(x) is unbounded when x = pi / 2. Therefore tan(x + pi/2) is unbounded at x = 0 and is therefore discontinuous.

Answer 2

y=1/x

or

y=(sin x)/x