Theorems, related to the continuity of functions and their uses in calculus, are presented and dicussed with examples.
Theorem 1
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:
lim_{x→0} sin(x)
lim_{x→ pi} cos(x)
lim_{x→ 1} arctan(x)
Solutions
Since sin(x) is continuous lim_{x→ 0} sin(x) = sin (0) = 0
Since cos(x) is continuous lim_{x→ pi} cos(x) = cos (pi) = 1
arctan(x) is continuous lim_{xx→ 1} arctan(x) = arctan (1) =  pi / 4
Theorem 2
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?
Solutions:
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.
Theorem 3
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) ]
Solutions:
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.
Theorem 4
If lim_{x→ a}g(x) = L and if f is a continuous function at L, then lim_{x→ a}f(g(x)) = f(lim_{x→ a}g(x)) = f(L).
Example:Evaluate the limit
lim_{xx→ a} sin (2x + 5)
Solution: sin x is continuous everywhere. Hence
lim_{x� a} sin (2x + 5) = sin (lim_{x� a} 2x + 5) = sin (2a + 5), 2x + 5 is a polynomial and therefore continuous everywhere.
Theorem 5
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.
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