Questions with answers on the continuity of functions with emphasis on piecewise functions.

__Example 1:__ For what values of x are each of the following functions discontinuous?

__Solution to Example 1__

a) For *x = 0*, the denominator of function *f(x)* is equal to *0* and *f(x)* is not defined and does not have a limit at *x = 0*. Therefore function *f(x)* is discontinuous at *x = 0*.

b) For *x = 2* the denominator of function *g(x)* is equal to 0 and function *g(x)* not defined at *x = 2* and it has no limit. Function *g(x)* is not continuous at *x = 2*.

c) The denominator of function *h(x)* can be factored as follows: *x*^{2} -1 = (x - 1)(x + 1). The denominator is equal to 0 for x = 1 and x = -1 values for which the function is undefined and has no limits. Function *h* is discontinuous at x = 1 and x = -1.

d) *tan(x)* is undefined for all values of *x* such that *x = π/2 + k π , where k is any integer (k = 0, -1, 1, -2, 2,...)* and is therefore discontinuous for these same values of *x*.

e) The denominator of function *j(x)* is equal to 0 for *x* such that *cos(x) - 1 = 0* or *x = k (2 π)*, where *k* is any integer and therefore this function is undefined and therefore discontinuous for all these same values of *x*.

f) Function k(x) is defined as the ratio of two continuous functions (with denominator x^{2} + 5 never equal to 0), is defined for all real values of *x* and therefore has no point of discontinuity.

g) * l(x) = (x + 4)/(x + 4) = 1 *. Hence lim l(x) as x approaches * -4 = 1 = l(-4) *. Function l(x) is continuous for all real values of x and therefore has no point of discontinuity.

__Example 2:__ Find *b such that **f(x)* given below is continuous?

__Solution to Example 2__

For x > -1, f(x) = 2 x^{ 2} + b is a polynomial function and therefore continuous.

For x < -1, f(x)= -x^{ 3} is a polynomial function and therefore continuous.

For x = -1

f(-1) = 2(-1)^{ 2} + b = 2 + b

let us consider the left and right hand limits

L1 = lim _{x→ -1 -} f(x) = -(-1)^{ 3} = 1 (limit from left of -1)

L2 = lim _{x→ -1 +} f(x) = 2(-1)^{ 2} + b = 2 + b (limit from right of -1)

For function f to be continuous, we need to have

L1 = L2 = 2 + b

or 2 + b = 1 or b = -1.

__Example 3:__ Find *a* and *b* such that both *g(x)* given below and its first derivative are continuous?

__Solution to Example 3__

__Continuity of function g__

For x > 2, g(x) = a x^{ 2} + b is a polynomial function and therefore continuous.

For x < 2, g(x) = -2 x + 2 is a polynomial function and therefore continuous.

let

L1 = lim _{x→ 2 -} g(x) = a (2)^{ 2} + b = 4 a + b

L2 = lim _{x→ 2 +} g(x) = -2(2) + 2 = -2

For continuity of g at x = 2, we need to have

L1 = L2 = g(2)

Which gives

4 a + b = -2

__Continuity of the derivative g'__

For x > 2, g '(x) = 2 a x is a polynomial function and therefore continuous.

For x < 2, g '(x) = -2 is a constant function and therefore continuous.

Let

l1 = lim _{x→ 2 -} g'(x) = 2 a (2) = 4 a

l2 = lim _{x→ 2 +} g'(x) = - 2

For continuity of g' at x = 2, we need to have

l1 = l2 or 4 a = - 2

The last equation gives: a = - 1 / 2. And substitute a by - 1 / 2 in the equation 4 a + b = -2 obtained above, we obtain b = 0.

More on Continuous Functions in Calculus

and
Continuity Theorems and Their use in Calculus