The Discontinuities of Some Common Functions
Polynomials: None. Every polynomial is continuous everywhere on
(− ∞, ∞).
Exponential functions (ex, 2x, 10x, etc.): None. Every exponential function
is continuous everywhere on (− ∞, ∞).
Trigonometric functions: For sin x and cos x, none, as both are continuous
everywhere on (− ∞, ∞). For the others, each has infinite discontinuities
wherever the denominator is 0 (infinitely many of them). That is, at ±π/2,
±3π/2, ±5π/2, ±7π/2, … , for tan x and sec x; and at 0, ±π, ±2π, ±3π, … , for
cot x and csc x.
Root functions ( n x ): None. Every n-th root function is continuous
everywhere on its domain, which is (− ∞, ∞) if n is odd, or [0, ∞) if n is even.
Rational functions: Various, from 0 to many, located at the zero(s) of the
denominator. Every discontinuity of a rational function is either a
removable discontinuity or an infinite discontinuity. It is NEVER a jump
discontinuity.
How to classify them? Simplify the rational function to its simplest
form by canceling all common factor(s) from the numerator and
denominator. For each zero of the denominator a, if the factor (x – a)
can be cancelled completely (thus, a is no longer a zero of the
denominator in the simplified expression), then the point at x = a on
the curve is a removable discontinuity. Otherwise, in the case that
(x – a) remains a factor of the denominator after simplification, the
point at x = a on the curve is an infinite discontinuity (it is also the
location of a vertical asymptote).
x2  4
Example 1: f ( x)  3
x  2x 2
The denominator has zeros at x = 0 and x = –2. Simplifying
x2  4
( x  2)( x  2) x  2
f ( x)  3

 2
2
x  2x
x 2 ( x  2)
x
In the simplified form, x = –2 is no longer a zero of the denominator, while
x = 0 still is a zero. Therefore, f has a removable discontinuity at x = –2,
and an infinite discontinuity at x = 0. (The graph has a vertical asymptote at
x = 0, but NOT at x = –2.)
Example 2: g ( x) 
x3  4x
x3  2x 2
Again, the denominator has zeros at x = 0 and x = –2. Simplifying
g ( x) 
x3  4x
x( x  2)( x  2) x  2


x3  2x 2
x 2 ( x  2)
x
As in the previous example, x = –2 is no longer a zero of the denominator,
and it is, therefore, a removable discontinuity. On the other hand, even
though there has been a cancellation of a common factor x from both
numerator and denominator, it was not a complete cancellation because there
remains another factor x in the denominator. Thus, x = 0 remains a zero of
the denominator and is an infinite discontinuity. (As before, the graph has a
vertical asymptote at x = 0.)
Example 3: h( x) 
x 2 ( x  1) 2 ( x  3)
x 2 ( x  1)( x  3) 2 ( x  7)
The zeros of denominators, and therefore discontinuities of h, are at x = –1,
0, 3, and 7. The function can be simplified as:
x 2 ( x  1) 2 ( x  3)
x 1
h( x )  2

2
x ( x  1)( x  3) ( x  7) ( x  3)( x  7)
The zeros of the denominator at x = –1 and x = 0 are completely cancelled,
so they are both removable discontinuities. The zeros at x = 3 (after a partial
cancellation of one of two copies) and at x =7 remain. Therefore, x = 3 and
x =7 are both infinite discontinuities of h, as well as the locations of vertical
asymptotes on the graph of h.