How to Classify Discontinuities:
Problems

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

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

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

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

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

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

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.

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

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.

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

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

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 $$x = 9$$.

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.

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 $$x = \frac 1 2$$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $$x = 4$$.

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

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 $$x = 1$$.

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

The limit from the left is infinite.

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

The limit from the right is infinite.

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

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

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

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

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

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

$$ f(4) = 4^2 - 9 = 7 $$

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

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

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

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

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

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

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.

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

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

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

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

This tells us the function is undefined 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:

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

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

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

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

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

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 $$x$$.)

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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