Theorems, related to the continuity of functions and their uses in calculus, are presented and dicussed with examples.
All polynomial functions and the functions sin x, cos x, arctan x and e x are continuous on the interval (-infinity , +infinity).
Example:Evaluate the following limits:
limx→ pi cos(x)
limx→ -1 arctan(x)
Since sin(x) is continuous limx→ 0 sin(x) = sin (0) = 0
Since cos(x) is continuous limx→ pi cos(x) = cos (pi) = -1
arctan(x) is continuous limxx→ -1 arctan(x) = arctan (-1) = - pi / 4
If functions f and g are continuous at x = a, then
A. (f + g) is continuous at x = a,
B. (f - g) is continuous at x = a,
C. (f . g) is continuous at x = a,
D. (f / g) is continuous at x = a if g(a) is not equal to zero.
If g(a) = 0 then (f / g) is discontinuous at x = a.
Example:Let f(x) = sin x and g(x) = cos x. Where are the following functions (f + g), (f - g), (f . g) and (f / g) continuous?
Since both sin x and cos x are continuous everywhere, according to theorem 2 above (f + g), (f - g), (f . g) are continuous everywhere.
However (f / g) is continuous everywhere except at values of x for which the denominator g(x) is equal to zero. These values are found by solving the trigonometric equation:
cos x = 0
The values which make cos x = 0 are given by:
x = pi/2 + k(pi) , where k is any integer.
(f / g) is continuous everywhere except at x = pi/2 + k(pi) , k integer.
A rational function is continuous everywhere except at the values of x that make the denominator of the function equal to zero.
Example:Find the values of x at which function f is discontinuous.
f(x) = (x - 2) / [ (2 x 2 + 2x - 4)(x 4 + 5) ]
The denominator of f is the product of two terms and is given by
(2 x 2 + 2x - 4)(x 4 + 5)
The term x 4 + 5 is always positive hence never equal to zero. We now need to find the zeros of 2 x 2 + 2x - 4 by solving the equation:
2 x 2 + 2x - 4 = 0
The solutions are: x = 1 and x = - 2
function f is discontinuous at x = 1 and x = -2.
If limx→ ag(x) = L and if f is a continuous function at L, then limx→ af(g(x)) = f(limx→ ag(x)) = f(L).
Example:Evaluate the limit
limxx→ a sin (2x + 5)
Solution: sin x is continuous everywhere. Hence
limx� a sin (2x + 5) = sin (limx� a 2x + 5) = sin (2a + 5), 2x + 5 is a polynomial and therefore continuous everywhere.
If g is a continuous function at x = a and function f is continuous at g(a), then the composition f o g is continuous at x = a.
Example:Show that any function of the form e ax + b is continuous everywhere, a and b real numbers.
f(x) = e x the exponential function and g(x) = ax + b a polynomial (linear) function are continuous everywhere. Hence the composition f(g(x)) = e ax + b is also continuous everywhere.
More on limits
Questions on Continuity with Solutions
Calculus Tutorials and Problems