Determine if Continuous Function if Not Classify Discotinuity

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Problem 1

Classify the discontinuity at $$x = -4$$ in the graph above.

Step 1

Examine the one-sided limits.

  • As $$x \to -4$$ from the left, we can see the function is heading downward. The arrow tells us the function is going to continue going downward infinitely.
  • As $$x \to -4$$ from the right, the same thing is happening, but in the positive direction.

Answer

The function appears to have an infinite discontinuity at x = -4.

Problem 2

Classify the discontinuity at $$x = -1$$ in the graph above.

Step 1

  • As $$x \to -1$$ from the left, we can see the function is heading to roughly $$y = \frac 1 2$$.
  • As $$x \to -1$$ from the right, the same thing is happening. The function appears to be headed towards $$y = \frac 1 2$$.
  • At $$x = -1$$ the function is defined and has a value of $$f(-1) = 2$$.

Answer

The function has a removable discontinuity at $$x = -1$$.

Problem 3

Classify the discontinuity at $$x = 2$$ in the graph above.

Step 1

  • As $$x\to 2$$ from the left, we can see the function is heading to roughly $$y = 3.5$$.
  • As $$x \to 2$$ from the right, the function appears to be headed towards $$y = 1$$.
  • At $$x = 2$$ the function is undefined.

Answer

The function appears to have a jump discontinuity at $$x = 2$$.

Problem 4

Classify the discontinuity at $$x = -3$$ in the graph above.

Step 1

  • As $$x\to -3$$ from the left, we can see the function seems to be heading to $$y = 1$$.
  • As $$x \to -3$$ from the right, the same thing is happening. The function appears to be headed towards $$y = 1$$.
  • At $$x = -3$$ the function is undefined.

Answer

The function appears to have a removable discontinuity at $$x = -3$$.

Problem 5

Classify the discontinuity at $$x = 1$$ in the graph above.

Step 1

  • As $$x \to 1$$ from the left, we can see the function seems to be heading to $$y = 1$$.
  • As $$x \to 1$$ from the right, the function grows infinitely large.
  • At $$x = 1$$ the function has a value of $$f(1) = 1$$.

Answer

The function appears to have a mixed discontinuity at $$x = 1$$.

Problem 6

Classify the discontinuity at $$x = 5$$ in the graph above.

Step 1

  • As $$x\to 5$$ from the left, we can see the function seems to be heading to about $$y = \frac 1 2$$
  • .
  • According to the graph, the function is not defined to the right of $$x = 5$$
  • .

Answer

The function appears to have an endpoint discontinuity at $$x=5$$.

Problem 7

Use the tables below to classify the discontinuity in the function at $$x = 8$$.

$$ \begin{array}{c|lcc|l} {x} & {f(x)}\\ \hline 7.9 & 1.36\\ 7.99 & 1.306\\ 7.999 & 1.3006\\ 7.9999 & 1.30006\\ 7.99999 & 1.300006 \end{array} $$

$$ \begin{array}{c|lcc|l} {x} & {f(x)}\\ \hline 8.1 & 4\\ 8.01 & 4.9\\ 8.001 & 4.95\\ 8.0001& 4.995\\ 8.00001 & 4.9995 \end{array} $$

Step 1

From the left: As $$x\to 8^-$$, the left-hand table implies the function is approaching 1.3.

From the right: As $$x\to 8^+$$, the right-hand table implies the function is approaching 5.

Answer

Since the two one-sided limits appear to be different, we conclude the function likely has a jump discontinuity at $$x=8$$.

$$ \begin{array}{c|lcc|l} {x} & {f(x)} \\ \hline 7.9 & 1.36\\ 7.99 & 1.306\\ 7.999 & 1.3006\\ 7.9999 & 1.30006\\ 7.99999 & 1.300006 \end{array} $$

$$ \begin{array}{c|lcc|l} {x} & {f(x)}\\ \hline 8.1 & 4\\ 8.01 & 4.9\\ 8.001 & 4.95\\ 8.0001& 4.995\\ 8.00001 & 4.9995 \end{array} $$

Problem 8

Suppose $$f(-6)=5.9$$. Use the tables below to classify the discontinuity in the function at $$x = -6$$.

$$ \begin{array}{c|lcc|l} {x} &{f(x)}\\ \hline -6.5 & 5.8\\ -6.05 & 5.89\\ -6.005 & 5.899\\ -6.0005 & 5.8999\\ -6.00005 & 5.89999 \end{array} $$

$$ \begin{array}{c|lcc|l} {x} &{f(x)}\\ \hline -5.9 & -2.1\\ -5.99 & -2.01\\ -5.999 & -2.001\\ -5.9999& -2.0001\\ -5.99999 & -2.00001 \end{array} $$

Step 1

From the left: As $$x \to -6^-$$, the table on the left implies $$f(x)$$ is approaching 5.9.

From the right: As $$x\to -6^+$$, the table on the right implies $$f(x)$$ is approaching -2.

Answer

The tables imply the one-sided limits have different values. So we conclude there is likely a jump discontinuity at $$x = -6$$.

$$ \begin{array}{c|lcc|l} {x} & {f(x)} \\ \hline -6.5 & 5.8\\ -6.05 & 5.89\\ -6.005 & 5.899\\ -6.0005 & 5.8999\\ -6.00005 & 5.89999 \end{array} $$

$$ \begin{array}{c|lcc|l} {x} &{f(x)}\\ \hline -5.9 & -2.1\\ -5.99 & -2.01\\ -5.999 & -2.001\\ -5.9999& -2.0001\\ -5.99999 & -2.00001 \end{array} $$

Problem 9

Suppose $$f(9)$$ is undefined. Use the tables below to classify the discontinuity in the function at there.

$$ \begin{array}{c|lcc|l} {x} &{f(x)}\\ \hline 8.9 & 24.3\\ 8.99 & 24.03\\ 8.999 & 24.003\\ 8.9999 & 24.0003\\ 8.99999 & 24.00003 \end{array} $$

$$ \begin{array}{c|lcc|l} {x} &{f(x)}\\ \hline 9.1 & 24.6\\ 9.01 & 24.06\\ 9.001 & 24.006\\ 9.0001 & 24.0006\\ 9.00001 & 24.00006 \end{array} $$

Step 1

Left-hand limit: As $$x$$ approaches 9 from the left, the tables imply $$f(x)$$ is approaching 24.

Right-hand limit: As $$x$$ approaches 9 from the right, the tables imply $$f(x)$$ approaches 24

Answer

The tables imply the limit exists as $$x$$ approaches 9, however the function is undefined there. We conclude the function likely has a removable discontinuity at $$x = 9$$.

$$ \begin{array}{c|lcc|l} {x} & {f(x)} \\ \hline 8.9 & 24.3\\ 8.99 & 24.03\\ 8.999 & 24.003\\ 8.9999 & 24.0003\\ 8.99999 & 24.00003 \end{array} $$

$$ \begin{array}{c|lcc|l} {x} & {f(x)}\\ \hline 9.1 & 24.6\\ 9.01 & 24.06\\ 9.001 & 24.006\\ 9.0001 & 24.0006\\ 9.00001 & 24.00006 \end{array} $$

Problem 10

Suppose $$f(1/2)=4$$. Use the tables below to classify the discontinuity in the function at there.

$$ \begin{array}{c|lcc|l} {x} &{f(x)}\\ \hline 0.40 & 6.9 \\ 0.49 & 6.99\\ 0.499 & 6.999\\ 0.4999 & 6.9999\\ 0.49999 & 6.99999 \end{array} $$

$$ \begin{array}{c|lcc|l} {x} &{f(x)}\\ \hline 0.6 & 7.25\\ 0.55 & 7.025\\ 0.505 & 7.0025\\ 0.5005 & 7.00025\\ 0.50005 & 7.000025 \end{array} $$

Step 1

Left-hand limit: As $$x$$ approaches $$\frac 1 2$$ from the left, the tables imply $$f(x)$$ approaches 7.

Right-hand limit: As $$x$$ approaches $$\frac 1 2$$ from the right, the tables imply $$f(x)$$ approaches 7.

Answer

The tables imply the limit exists and is equal to 7, but we know the function value is $$f(1/2) = 4$$. Since the limit and the function have different values, we conclude the function has a removable discontinuity at $$x = \frac 1 2$$.

$$ \begin{array}{c|lcc|l} {x} & {f(x)} \\ \hline 0.40 & 6.9\\ 0.49 & 6.99\\ 0.499 & 6.999\\ 0.4999 & 6.9999\\ 0.49999 & 6.99999 \end{array} $$

$$ \begin{array}{c|lcc|l} {x} & {f(x)}\\ \hline 0.6 & 7.25\\ 0.55 & 7.025\\ 0.505 & 7.0025\\ 0.5005 & 7.00025\\ 0.50005 & 7.000025 \end{array} $$

Problem 11

Classify the discontinuity in $$f(x)$$ at $$x = 11$$.

$$ f(x) = \left\{% \begin{array}{ll} 3x+1, & x\leq 11\\ 4x -2, & x > 11 \end{array} \right. $$

Step 1

Examine the limit from the left.

$$ \displaystyle\lim_{x\to 11^-} f(x) = \displaystyle\lim_{x\to 11^-} (3x + 1) = 3(11) + 1 = 34 $$

Step 2

Examine the limit from the right.

$$ \displaystyle\lim_{x\to 11^+} f(x) = \displaystyle\lim_{x\to 11^+} (4x-2) = 4(11) -2 = 44 - 2 = 42 $$

Answer

The function has a jump discontinuity at $$x = 11$$.

Problem 12

Is the function continuous at $$x = -2$$? If not, classify the type of discontinuity there.

$$ f(x) = \left\{% \begin{array}{ll} \sqrt{x+6}, & x\leq -2\\ -x^2, & x > -2 \end{array} \right. $$

Step 1

Examine the left-hand limit.

$$ \displaystyle\lim_{x\to-2^-} f(x) = \displaystyle\lim_{x\to-2^-} \sqrt{x+6} = \sqrt{-2 + 6} = \sqrt 4 = 2 $$

Step 2

Examine the right-hand limit.

$$ \displaystyle\lim_{x\to-2^+} f(x) = \displaystyle\lim_{x\to-2^+} -x^2 = -(-2)^2 = -4 $$

Answer

The one-sided limits are different, so the function has a jump discontinuity at $$x = -2$$.

Problem 13

Classify the discontinuity in $$f(x)$$ at $$x = 3$$.

$$ f(x) = \left\{% \begin{array}{ll} |x|, & x < 3\\ 4x+2, & x > 3 \end{array} \right. $$

Step 1

Examine the limit from the left.

$$ \displaystyle\lim_{x\to 3^-} f(x) = \displaystyle\lim_{x\to 3^-} |x| = |3| = 3 $$

Step 2

Examine the limit from the right.

$$ \displaystyle\lim_{x\to 3^-} f(x) = \displaystyle\lim_{x\to 3^-} (4x+2) = 4(3)+2 = 12 +2 = 14 $$

Answer

Since the one-sided limits are different, the function has a jump discontinuity at $$x = 3$$.

Problem 14

Is the function continuous at $$x = 0$$? If not, classify the type of discontinuity there.

$$ f(x) = \left\{% \begin{array}{ll} x^{1/3}, & x < 0\\ x^{1/2}, & x > 0 \end{array} \right. $$

Step 1

$$ \begin{align*} \lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} x^{1/3} = 0^{1/3} = 0\\[6pt] \lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} x^{1/2} = 0^{1/2} = 0 \end{align*} $$

Step 2

Examine the function value at $$x = 0$$.

$$ f(0)\mbox{ is undefined.} $$

Answer

The one-sided limits are the same, so the limit exists. But since the function is undefined, it has a removable discontinuity at $$x = 0$$.

Problem 15

Classify the discontinuity in $$f(x)$$ at $$x = 4$$.

$$ f(x) = \left\{% \begin{array}{ll} \sin\left(\frac{\pi} 8 x\right), & x < 4\\ 2, & x = 4\\ \cos(x-4), & x > 4 \end{array} \right. $$

Step 1

$$ \begin{align*} \lim_{x\to 4^-} f(x) & = \lim_{x\to 4^-} \sin\left(\frac \pi 8 x\right) = \sin\frac \pi 2 = 1\\[6pt] \lim_{x\to 4^+} f(x) & = \lim_{x\to 4^+} \cos(x-4) =\cos(4-4) = \cos 0 = 1 \end{align*} $$

Answer

The limit value is 1, but the function value is 2. Since the limit and the function have different values, the function has a removable discontinuity at $$x = 4$$.

Problem 16

Is the function continuous at $$x = 1$$? If not, classify the type of discontinuity there.

$$ f(x) = \left\{% \begin{array}{ll} e^{x-1}, & x < 1\\ 0, & x = 1\\ -\cos\left(\pi x\right), & x > 1 \end{array} \right. $$

Step 1

$$ \begin{align*} \lim_{x\to 1^-} f(x) & = \lim_{x\to 1^-} e^{x-1} = e^{1-1} = e^0 = 1\\[6pt] \lim_{x\to 1^-} f(x) & = \lim_{x\to 1^-} -\cos(\pi x) = -\cos \pi = -(-1) = 1 \end{align*} $$

Answer

The limit value is 1, but the function value is $$f(1) = 0$$. Since the limit and function values are different, there is a removable discontinuity at $$x = 1$$.

Problem 17

Classify the discontinuity in $$f(x)$$ at $$x = 1$$.

$$ f(x) = \left\{% \begin{array}{ll} \frac 1 {x-1}, & x < 1\\[6pt] \frac 1 {x^2 - 2x + 1}, & x > 1 \end{array} \right. $$

Step 1

Examine the limit from the left.

  • The limit has the form of $$\lim\limits_{x\to 1^-} f(x) = \frac 1 0$$.

The limit from the left is infinite.

Step 2

Examine the limit from the right.

  • The limit has the form of $$\lim\limits_{x\to 1^+} f(x) = \frac 1 0$$.

The limit from the right is infinite.

Answer

Since both one-sided limits are infinite, the function has an infinite discontinuity at $$x = 1$$.

Problem 18

Is the function continuous at $$x = 0$$? If not, classify the type of discontinuity there.

$$ f(x) = \left\{% \begin{array}{ll} \cot x, & x < 0\\[6pt] \ln x, & x > 0 \end{array} \right. $$

Step 1

Examine the limit from the left.

$$ \begin{align*} \lim_{x\to0^-} f(x) & = \lim_{x\to0^-} \cot x\\[6pt] & = \lim_{x\to0^-} \frac{\cos x}{\sin x}\\[6pt] & = \frac{\cos 0}{\sin 0}\\[6pt] & = \frac 1 0 \end{align*} $$

This tells us $$\lim\limits_{x\to0^-} f(x)$$ is infinite.

Step 2

Examine the limit from the right.

$$ \displaystyle\lim_{x\to0^+} f(x) = \displaystyle\lim_{x\to0^+} \ln x = -\infty $$

Answer

Since both one-sided limits are infinite, the function has an infinite discontinuity at $$x = 0$$.

Problem 19

Is the function continuous at $$x = 4$$? If not, classify the type of discontinuity there.

$$ f(x) = \left\{% \begin{array}{ll} x+3, & x < 4\\[6pt] x^2 - 9, & x \geq 4 \end{array} \right. $$

Step 1

$$ \begin{align*} \lim_{x\to 4^-} f(x) & \lim_{x\to4^-} (x+3) = 4 + 3 = 7\\[6pt] \lim_{x\to 4^+} f(x) & \lim_{x\to4^+} (x^2-9) = (4)^2 - 9 = 7\ \end{align*} $$

Answer

Since the limit value and function value are the same, the function is continuous at $$x = 4$$.

Problem 20

Is the function continuous at $$x = -1$$? If not, classify the type of discontinuity there.

$$ f(x) = \left\{% \begin{array}{ll} 2x + 5, & x < -1\\[6pt] 2-x, & x \geq -1 \end{array} \right. $$

Step 1

$$ \begin{align*} \lim_{x\to-1^-} f(x) & = \lim_{x\to-1^-} (2x+5) = 2(-1) + 5 = 3\\[6pt] \lim_{x\to-1^+} f(x) & = \lim_{x\to-1^+} (2 - x) = 2 - (-1) = 3 \end{align*} $$

Step 2

Examine the function value at $$x = -1$$.

$$ f(-1) = 2 - (-1) = 3 $$

Answer

Since the limit values and function value are the same, the function is continuous at $$x = -1$$.

Problem 21

Determine where the function is discontinuous, and classify the discontinuities.

$$ f(x) = \frac{x^2 + 5x - 6}{x + 6} $$

Step 1

Examine the function for places where it will be undefined. This is similar to finding the domain of the function.

Since the function is rational, it will be undefined where the denominator is zero. In this case, at $$x = -6$$.

Step 2

Determine the form of the limit as $$x$$ approaches -6.

$$ \lim_{x\to-6} f(x) = \frac{(-6)^2 + 5(-6) - 6}{(-6) + 6} = \frac{36 -30 - 6}{-6 + 6} = \frac{0}{0} $$

With the $$\frac 0 0$$ form this function either has a removable discontinuity (if the limit exists) or an infinite discontinuity (if the one-sided limits are infinite) at -6.

Step 3

Find and divide out any common factors.

$$ \frac{x^2 + 5x - 6}{x + 6} = \frac{\blue{(x+6)}(x-1)}{\blue{x + 6}} = x-1 $$

Step 4

Evaluate the limit of this simpler function.

$$ \displaystyle\lim_{x\to-6} (x-1) = -6-1 = -7 $$

Answer

Since the limit at $$x = -6$$ exists, but the function does not, the function has a removable discontinuity there.

Problem 22

What type of discontinuity is does the function exhibit at $$x = 0$$?

$$ f(x) = \frac{\sin 4x}{2x^2+5x} $$

Step 1

Determine $$f(0)$$

$$ f(0) = \frac{\sin 0}{2(0)^2 + 5(0)} = \frac 0 0 $$

This tells us the function is undefined at $$x = 0$$.

Step 2

Determine whether or not the limit exists at $$x = 0$$.

Examine the limit as $$x$$ approaches 0.

$$ \lim_{x\to 0} f(x) = \frac{\sin 0}{2(0)^2 + 5(0)} = \frac 0 0 $$

With this form, we know the limit might exist. Using the techniques from earlier lessons (see Indeterminate Limits---Sine Forms), we evaluate the limit as follows:

$$ \begin{align*} \lim_{x\to 0} f(x) & = \lim_{x\to0} \frac{\sin 4x}{2x^2 + 5x}\\[6pt] & = \lim_{x\to0} \frac{\sin 4x}{x(2x + 5)}\\[6pt] & = \lim_{x\to0} \frac{\sin 4x} x \cdot \frac 1 {2x + 5}\\[6pt] & = \lim_{x\to0} \frac{\blue 4}{\blue 4}\cdot\frac{\sin 4x} x \cdot \frac 1 {2x + 5}\\[6pt] & = \lim_{x\to0} \frac{\blue 4} 1\cdot \frac{\sin 4x}{\blue 4 x} \cdot \frac 1 {2x + 5}\\[6pt] & = \blue 4 \cdot\red{\lim_{x\to0} \frac{\sin 4x}{4x}} \cdot \frac 1 {2x + 5}\\[6pt] & = \blue 4 \cdot \red{1} \cdot \frac 1 {2(0) + 5}\\[6pt] & = \frac 4 5 \end{align*} $$

Answer

At $$x = 0$$, the limit exists, but the function does not. This means the function has a removable discontinuity at $$x = 0$$.

Problem 23

Determine where the function is discontinuous, and classify the discontinuities.

$$ f(x) = \frac{2x-3}{x+7} $$

Step 1

Identify the points where the function is undefined.

This function is undefined at $$x = -7$$.

Step 2

Examine the form of the limit at $$x = -7$$.

$$ \displaystyle\lim_{x\to-7} f(x) = \frac{2(-7) -3}{0} = \frac{-17} 0 $$

Answer

The $$\frac n 0$$ form indicates the function has an infinite discontinuity at $$x = -7$$.

Problem 24

Determine where the function is discontinuous for $$0\leq x \leq \frac \pi 2$$, and classify the discontinuities.

$$ f(x) = \frac{x^2+9x}{1-\cos 3x} $$

Step 1

Determine where the function is undefined.

Since we are restricted to $$0\leq x \leq \frac \pi 2$$, we can determine that the denominator is equal to 0 only when

$$ \begin{align*} 1 - \cos 3x & = 0\\ \cos 3x & = 1\\ 3x & = 0\\ x & = 0\\ \end{align*} $$

(The denominator is also equal to zero when $$x = \frac{2n\pi} 3$$ for integer values of $$n$$, but these are outside the bounds set for $$x$$.)

Step 2

Determine the form of the limit from the right at $$x=0$$ (note that the limit from the left doesn't exist since that is outside the bounds allowed for $$x$$).

$$ \lim_{x\to0^+} f(x) = \frac{(0)^2 + 9(0)}{1 - \cos 0} = \frac 0 0 $$

Step 3

$$ \begin{align*} \lim_{x\to0^+} \frac{x^2 + 9x}{1-\cos 3x} & = \lim_{x\to0^+} \frac{x(x + 9)}{1-\cos x3x}\\[6pt] & = \lim_{x\to0^+} \frac x {1-\cos 3x}\cdot \frac{x + 9} 1\\[6pt] & = \lim_{x\to0^+} \frac{\blue 3}{\blue 3}\cdot \frac x {1-\cos 3x}\cdot \frac{x + 9} 1\\[6pt] & = \lim_{x\to0^+} \frac 1 {\blue 3}\cdot \frac{\blue 3 x} {1-\cos 3x}\cdot \frac{x + 9} 1\\[6pt] & = \lim_{x\to0^+} \frac{\blue 3 x} {1-\cos 3x}\cdot \frac{x + 9} {\blue 3}\\[6pt] & = \lim_{x\to0^+} \red{\frac{3x} {1-\cos 3x}}\cdot \frac{x + 9} 3\\[6pt] & = \red{\frac 1 0}\cdot \frac{0 + 9} 3\\[6pt] & = \frac 9 0 \end{align*} $$

(Since $$\lim_{\theta\to 0} \frac{1 - \cos \theta}{\theta} = 0 = \frac 0 1$$, we know $$\lim_{\theta\to 0} \red{\frac{\theta}{1 - \cos \theta}} = \red{\frac 1 0}$$)

Answer

The limit has an infinite discontinuity at $$x = 0$$.

Problem 25

Determine where the function is discontinuous, and classify the discontinuities.

$$ f(x) = \frac{\sqrt x}{x^2+1} $$

Step 1

What is the domain of the function?

The function is defined at all $$x\geq 0$$.

Step 2

Determine the limit as $$x$$ approaches 0 from the right.

$$ \displaystyle\lim_{x\to 0^+}\frac{\sqrt x}{x^2+1} = \frac{ \sqrt 0} 1 = 0 $$

Step 3

Determine the value of $$f(0)$$.

$$ f(0) = \frac{\sqrt 0}{0+1} = 0 $$

Answer

At $$x=0$$, the limit from the right exists and is equal to the function value. Consequently, the function has an endpoint discontinuity at $$x=0$$. We could also say the function is continuous from the right.

Problem 26

Determine where the function is discontinuous, and classify the discontinuities.

$$ f(x) =\left\{% \begin{array}{ll} |x+2|, & 0\leq x < 4\\ 6-(x-4)^2, & 4 \leq x < 8\\ \end{array} \right. $$

Step 1

At $$x = 0$$

  • $$f(0) = |0 + 2| = 2$$
  • $$\lim\limits_{x\to0^+} f(x) = |0+2| = 2$$

Since the function and the limit value are the same, the function has an end-point discontinuity at $$x = 0$$.

Step 2

At $$x = 8$$

  • $$\lim\limits_{x\to8^-} f(x) = 6 - (8 - 4)^2 = 6 - 16 = -10$$

Since the limit exists at $$x = 8$$ we could call this a removable discontinuity which, if fixed, would still be an endpoint discontinuity.

Step 3

At $$x = 4$$

  • $$\displaystyle\lim_{x\to4^-} f(x) = |4+2| = 6$$
  • $$\displaystyle\lim_{x\to 4^+} f(x) = 6 - (4-4)^2 = 6$$
  • $$f(4) = 6 - (4-4)^2 = 6$$

Since $$f(4) = \lim\limits_{x\to4} f(x)$$, we know the function is continuous at $$x = 4$$.

Answer

The function has endpoint discontinuities at $$x = 0$$ and $$x = 8$$.

Return to lesson

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