Questions with answers on the continuity of functions with emphasis on rational and piecewise functions. The continuity of a function and its derivative at a given point is discussed. Graphical meaning and interpretation of continuity are also included.

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

a) \;\; f(x) = \dfrac{1}{x} \;\; b) \;\; g(x) = \dfrac{2}{x-2} \;\; c) \;\; h(x) = \dfrac{x+1}{x^2-1}

d) \;\; i(x) = \tan(x) \;\; e) \;\; j(x) = \dfrac{1}{\cos(x)-1} \;\; f) \;\; k(x) = \dfrac{x+2}{x^2+5}

g) \;\; l(x) = \begin{cases}
\dfrac{x+4}{x+4} & x \ne -4 \\
1 & x = -4 \\
\end{cases}

__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

d)

e) The denominator of function

f) Function k(x) is defined as the ratio of two continuous functions (with denominator x

g)

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

f(x) = \begin{cases}
2x^2+b & x \ge -1 \\
-x^3 & x \textless -1 \\
\end{cases}

__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

limit from left of -1

L1 = \lim_{x\to\ -1^-} f(x) = -(-1)^3 = 1

limit from right of -1

L2 = \lim_{x\to\ -1^+} f(x) = 2(-1)^2 + b = 2 + b

For function f to be continuous, we need to have

L1 = L2 = 2 + b

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

Substitute b by -1 in the given function to obtain

f(x) = \begin{cases}
2x^2-1 & x \ge -1 \\
-x^3 & x \textless -1 \\
\end{cases}

The graph of f is shown below and it is clear that the function is continuous at x = -1.
Example 3: Find *a* and *b* such that both *g(x)* given below and its first derivative are continuous?

g(x) = \begin{cases}
ax^2+b & x \ge 2 \\
-2x+2 & x \textless 2 \\
\end{cases}

__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\to\ 2^+} g(x) = a (2)^2 + b = 4 a + b

L2 = \lim_{x\to\ 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

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\to\ 2^+} g'(x) = 2a(2) = 4 a

l2 = \lim_{x\to\ 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. Substitute a and b by their values to obtain:

g(x) = \begin{cases}
-\dfrac{1}{2}x^2 & x \ge 2 \\
-2x+2 & x \textless 2 \\
\end{cases}

Function g(x) is graphed below and it is clear that both the function and its derivative (slope) are continuous at x = 2.
More on Continuous Functions in Calculus

and Continuity Theorems and Their use in Calculus