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.

Examples with Solutions

Example 1:

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 $x = \frac{\pi}{2} + k \pi ,$ 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 $x = k (2 \pi)$, 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) $l(x) = \dfrac{x + 4}{x + 4} = 1$ for $x \ne - 4$.
$\lim_{x \to -4} l(x) = 1 = l(-4)$.
Function $l(x)$ is continuous for all real values of $x$ and therefore has no point of discontinuity.

Example 2:

Solution to Example 2
For $x > -1$, $f(x) = 2 x^ 2 + b$ is a polynomial function and therefore continuous.
For $x \lt -1$, $f(x) = -x^3$ 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
$L1 = \lim_{x\to -1^-} f(x) = -(-1)^3 = 1$
$L2 = \lim_{x\to -1^+} f(x) = 2(-1)^2 + b = 2 + b$
For function $f$ to be continuous, we need to have
$L2 = L1$
or $2 + b = 1$
Solve for $b$
$b = -1$.
Substitute $b$ by -1 in the given function to obtain $$f(x) = \begin{cases} 2x^2-1 & x \ge -1 \\ -x^3 & x \lt -1 \\ \end{cases}$$ The graph of $f$ is shown below and it is clear that the function is continuous at $x = -1$.

Example 3:

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
$4 a + b = -2$
Continuity of the derivative $g'$
For $x > 2$, $g '(x) = 2 a x$ is a polynomial function and therefore continuous.
For $x \lt 2$, $g '(x) = -2$ 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 $4 a = - 2$
Solve the last equation to obtain: $a = - 1 / 2$.
Substitute $a$ by $- 1 / 2$ in the equation $4 a + b = -2$ obtained above and solve for $b$ to obtain $b = 0$.
Substitute $a$ and $b$ by their values to obtain function $g$: $$g(x) = \begin{cases} -\dfrac{1}{2}x^2 & x \ge 2 \\ -2x+2 & x \lt 2 \\ \end{cases}$$ Function $g(x)$ is graphed below and it is clear that both the function and its derivative (slope) are continuous at $x = 2$.

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