Continuity Theorems and Their Applications in Calculus
Theorems, related to the continuity of functions and their applications in calculus are presented and discussed with examples.
Theorem 1All 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\to\0} \sin (x)
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\lim_{x\to\pi} \cos (x)
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\lim_{x\to\ -1} \arctan(x)
Solutions If function f is continuous at x = a, then
\lim_{x\to\ a} \f(x) = f(a)
Since sin(x) is continuous
\lim_{x\to\0} \sin (x) = \sin(0) = 0
Since cos(x) is continuous
\lim_{x\to\pi} \cos (x) = \cos(\pi) = - 1
Since arctan(x) is continuous
\lim_{x\to\ -1} \arctan(x) = \arctan(-1) = - \pi / 4
Theorem 2If functions f and g are continuous at x = a, thenA. (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 make 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 = π/2 + k(π) , where k is any integer. (f / g) is continuous everywhere except at x = π/2 + k(π) , k integer.
Theorem 3A 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) = \dfrac{x-2}{(2 x^2 + 2 x - 4)(x^4 + 5)}
Solutions: The denominator of f is the product of two terms and is given by
(2 x^2 + 2 x - 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 + 2 x - 4 = 0
The solutions are: x = 1 and x = - 2 function f is discontinuous at x = 1 and x = -2.
Theorem 4
\color{red}{\lim_{x\to\ a} g(x) = L}
\color{red}{\lim_{x\to\ a} f(g(x)) = f(\lim_{x\to\ a} g(x)) = f(L)}
Example:Evaluate the limit
\lim_{x\to\ a} \sin(2x + 5)
Solution: sin x is continuous everywhere and 2 x + 5 is a polynomial and also continuous everywhere. Hence
\lim_{x\to\ a} \sin(2x + 5) = \sin(\lim_{x\to\ a}(2x+5)) = \sin(2a + 5)
Theorem 5If 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 Questions on Continuity with Solutions Calculus Tutorials and Problems |
