How To Find Discontinuity Of A Function

Quick Overview

On graphs, the open and airtight circles, or vertical asymptotes drawn as dashed lines help united states place discontinuities.
As before, graphs and tables allow u.s. to estimate at best.
When working with formulas, getting zero in the denominator indicates a point of aperture.
When working with piecewise-defined functions, cheque for discontinuities at the transition points where 1 piece ends and the next begins.

Examples

Case ane

Using the graph shown beneath, place and classify each point of discontinuity.

Step 1

The table below lists the location ($$ten$$-value) of each discontinuity, and the type of discontinuity.

$$ \brainstorm{array}{c|l} {10} & {\mbox{Type}}\\ \hline -seven & \mbox{Mixed}\\ -iii & \mbox{Removable}\\ ii & \mbox{Jump}\\ four & \mbox{Space}\\ half-dozen & \mbox{Endpoint} \terminate{assortment} $$

Note that the discontinuity at $$x=-7$$ is both removable (the part value is different from the ane-sided limit value) and an endpoint (since the graph is not defined to the left of $$ten=-7$$).

Instance ii

Using the tables below, what type of discontinuity seems to exist at $$10 = 5$$?

$$ \begin{array}{c|lcc|fifty} {x} & {f(x)}\\ \hline 4.9 & 8.15\\ 4.99 & eight.015\\ iv.999 & 8.0015\\ iv.9999 & 8.00015\\ iv.99999 & 8.000015\\ \finish{array} $$

$$ \begin{array}{c|lcc|50} {x} & {f(ten)}\\ \hline 5.1 & 2.4\\ v.01 & 2.43\\ 5.001 & 2.403\\ v.0001& 2.4003\\ 5.00001 & 2.40003 \end{assortment} $$

Stride 1

Examine the 1-sided limits.

The table on the left tells u.s. $$\lim\limits_{x\to5^-}f(x) \approx 8$$

The table on the right tells us $$\lim\limits_{10\to5^+}f(x) \approx two.4$$

Answer

The tables pb u.s. to believe the i-sided limits are different, so we conclude the function likely has a jump discontinuity at $$ten = 5$$.

Example 3

Is the role below continuous at its transition point? If not, identify the blazon of discontinuity occurring there.

$$ f(x) = \left\{% \begin{assortment}{ll} x^two, & x\leq one\\ x+three, & x > one \end{array} \right. $$

Step i

Identify the transition point(s).

The transition betoken is at $$x = 1$$ since this is where the part transitions from one formula to the next.

Step 2

Decide the left-hand limit at the transition point.

$$ \displaystyle\lim_{x\to one^-} f(x) = \displaystyle\lim_{10\to ane^-} x^2 = ane^two = 1 $$

Step 3

Determine the right-paw limit at the transition point.

$$ \displaystyle\lim_{x\to i^+} f(10) = \displaystyle\lim_{ten\to 1^+} (ten+ iii) = one + iii = four $$

Answer

Since the one-sided limits are different, the role has a jump aperture at $$x = i$$.

Example iv

Is the part below continuous at x = four? If not, identify the type of discontinuity occurring there.

$$ f(ten) = \left\{% \brainstorm{array}{ll} \sqrt x, & 0\leq 10 < iv\\ 5,& x = 4\\ 6 - 10, & x > 4 \end{array} \correct. $$

Step one

Examine the left-paw limit.

$$ \displaystyle\lim_{ten\to 4^-} f(x) = \displaystyle\lim_{x\to 4^-} \sqrt x = \sqrt{four} = two $$

Stride 2

Examine the right-hand limit.

$$ \displaystyle\lim_{x\to four^+} f(x) = \displaystyle\lim_{x\to iv^+} (6-10) = 6 -4 = 2 $$

Step 3

Decide the office value.

$$ f(4) = v $$

Answer

The limit exists, and the part exists, but they have different values. The part has a removable aperture at $$x = 4$$.

Example 5

Without graphing, decide the type of discontinuity the function below has at $$ten = 3$$.

$$ f(x) = \frac{x^2 + 2x - 15}{10^two-2x-iii} $$

Footstep 1

Evaluate $$f(3)$$

$$ f(\blueish three) = \frac{(\blue{3})^ii + 2(\blue{3}) - 15}{(\blue{three})^2-2(\blueish{3})-3} = \frac{9 + 6 - 15}{9-6-three} = \frac 0 0 $$

The function is undefined at $$x = 3$$, so there is a discontinuity at this point. To determine the type, we will need to evaluate the limit as $$x$$ approaches three.

Pace 2

Since the function has a $$\frac 0 0$$ form at $$x = 3$$, nosotros need to notice and dissever out the common factors in the numerator and denominator.

$$ \frac{x^two + 2x - 15}{x^2-2x-3} = \frac{(x+v)\blue{(x-3)}}{\blue{(x-3)}(x+1)} = \frac{x+5}{x+1} $$

Step 3

Evaluate the limit of the simpler function as $$x$$ approaches 3.

$$ \displaystyle\lim_{\blue{x\to3}} \frac{x+5}{10+1} = \frac{\blue 3 + 5}{\blue 3 + 1} = \frac 8 four = 2 $$

Answer

Since the limit exists, but the function value does non, we know the function has is a removable discontinuity at $$x = 3$$.

Example 6

Without graphing, determine the blazon of aperture the role below has at $$x = -1$$.

$$ f(x) = \frac{x^two + 2x - 15}{x^2-2x-three} $$

Footstep one

Evaluate $$f(-1)$$

$$ f(\blue{-1}) = \frac{(\blue{-1})^2 + 2(\blue{-1}) - 15}{(\blue{-ane})^two-2(\blueish{-1})-3} = \frac{ane - 2 - fifteen}{1+two-3} = \frac{-16} 0 $$

Since we have segmentation by zero, the function doesn't exist at $$10 = -1$$. But, the $$\frac north 0$$ form tells us the role is becoming infinitely large as $$x$$ approaches -1.

Note: In club to make up one's mind if the limit is space, we would need to know which management the function is going equally $$10$$ approached -i. Only for the purposes of classifying the aperture, information technology'due south plenty to know the function becomes infinitely large.

Answer

The office has an infinite discontinuity at $$10 = -one$$.

Go along to Problems

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