# Questions on Continuity with Solutions

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?

__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

limit from left of -1

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.

Substitute b by -1 in the given function to obtain

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

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

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:

More on Continuous Functions in Calculus

and Continuity Theorems and Their use in Calculus