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Finding Limits: Properties of Limits

Finding Limits: Properties of Limits

By the end of this section, you will be able to:

  • Find the limit of a sum, a difference, and a product
  • Find the limit of a polynomial
  • Find the limit of a power or a root
  • Find the limit of a quotient

Consider the rational function

f(x)=x26x7x7f(x)=\frac{x^2-6x-7}{x-7}

The function can be factored as follows:

f(x)=(x7)(x+1)x7, which gives us f(x)=x+1, x7.f(x)=\frac{(x-7)(x+1)}{x-7},\text{ which gives us }f(x)=x+1,\ x\ne7.

Does this mean the function ff is the same as the function g(x)=x+1g(x)=x+1?

The answer is no. Function ff does not have x=7x=7 in its domain, but gg does. Graphically, we observe there is a hole in the graph of f(x)f(x) at x=7x=7, as shown below, and no such hole in the graph of g(x)g(x), as shown after it.

The graph of function ff contains a break at x=7x=7 and is therefore not continuous at x=7x=7.

The graph of function gg is continuous.

So, do these two different functions also have different limits as xx approaches 7?

Not necessarily. Remember, in determining a limit of a function as xx approaches aa, what matters is whether the output approaches a real number as we get close to x=ax=a. The existence of a limit does not depend on what happens when xx equals aa.

Look again at the two graphs above. Notice that in both graphs, as xx approaches 7, the output values approach 8. This means

limx7f(x)=limx7g(x).\lim\limits_{x\to7}f(x)=\lim\limits_{x\to7}g(x).

Remember that when determining a limit, the concern is what occurs near x=ax=a, not at x=ax=a. In this section, we will use a variety of methods, such as rewriting functions by factoring, to evaluate the limit. These methods will give us formal verification for what we formerly accomplished by intuition.

Finding the Limit of a Sum, a Difference, and a Product

Graphing a function or exploring a table of values to determine a limit can be cumbersome and time-consuming. When possible, it is more efficient to use the properties of limits, which is a collection of theorems for finding limits.

Knowing the properties of limits allows us to compute limits directly. We can add, subtract, multiply, and divide the limits of functions as if we were performing the operations on the functions themselves to find the limit of the result. Similarly, we can find the limit of a function raised to a power by raising the limit to that power. We can also find the limit of the root of a function by taking the root of the limit. Using these operations on limits, we can find the limits of more complex functions by finding the limits of their simpler component functions.

Properties of Limits. Let aa, kk, AA, and BB represent real numbers, and ff and gg be functions, such that limxaf(x)=A\lim_{x\to a}f(x)=A and limxag(x)=B\lim_{x\to a}g(x)=B. For limits that exist and are finite, the properties of limits are summarized below.

PropertyFormula
Constant, kklimxak=k\lim_{x\to a}k=k
Constant times a functionlimxa[kf(x)]=klimxaf(x)=kA\lim_{x\to a}[k\cdot f(x)]=k\lim_{x\to a}f(x)=kA
Sum of functionslimxa[f(x)+g(x)]=limxaf(x)+limxag(x)=A+B\lim_{x\to a}[f(x)+g(x)]=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)=A+B
Difference of functionslimxa[f(x)g(x)]=limxaf(x)limxag(x)=AB\lim_{x\to a}[f(x)-g(x)]=\lim_{x\to a}f(x)-\lim_{x\to a}g(x)=A-B
Product of functionslimxa[f(x)g(x)]=limxaf(x)limxag(x)=AB\lim_{x\to a}[f(x)\cdot g(x)]=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)=A\cdot B
Quotient of functionslimxaf(x)g(x)=limxaf(x)limxag(x)=AB, B0\lim_{x\to a}\tfrac{f(x)}{g(x)}=\tfrac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}=\tfrac{A}{B},\ B\ne0
Function raised to an exponentlimxa[f(x)]n=[limxaf(x)]n=An\lim_{x\to a}[f(x)]^n=[\lim_{x\to a}f(x)]^n=A^n, where nn is a positive integer
nnth root of a function, where nn is a positive integerlimxaf(x)n=limxa[f(x)]n=An\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x\to a}[f(x)]}=\sqrt[n]{A}
Polynomial functionlimxap(x)=p(a)\lim_{x\to a}p(x)=p(a)

Example. Evaluate limx3(2x+5)\lim_{x\to3}(2x+5).

Solution.

limx3(2x+5)=limx3(2x)+limx3(5)Sum of functions property=2limx3(x)+limx3(5)Constant times a function property=2(3)+5Evaluate=11 \begin{array}{ll} \lim\limits_{x\to3}(2x+5)=\lim\limits_{x\to3}(2x)+\lim\limits_{x\to3}(5) & \text{Sum of functions property} \\ =2\lim\limits_{x\to3}(x)+\lim\limits_{x\to3}(5) & \text{Constant times a function property} \\ =2(3)+5 & \text{Evaluate} \\ =11 & \end{array}

Evaluate the following limit:limx12(2x+2)\lim_{x\to-12}(-2x+2).

Finding the Limit of a Polynomial

Not all functions or their limits involve simple addition, subtraction, or multiplication. Some may include polynomials. Recall that a polynomial is an expression consisting of the sum of two or more terms, each of which consists of a constant and a variable raised to a nonnegative integral power. To find the limit of a polynomial function, we can find the limits of the individual terms of the function, and then add them together. Also, the limit of a polynomial function as xx approaches aa is equivalent to simply evaluating the function for aa.

How To: Given a function containing a polynomial, find its limit.

  1. Use the properties of limits to break up the polynomial into individual terms.
  2. Find the limits of the individual terms.
  3. Add the limits together.
  4. Alternatively, evaluate the function for aa.

Example. Evaluate limx3(5x2)\lim_{x\to3}(5x^2).

Solution.

limx3(5x2)=5limx3(x2)Constant times a function property=5(32)Function raised to an exponent property=45 \begin{array}{ll} \lim\limits_{x\to3}(5x^2)=5\lim\limits_{x\to3}(x^2) & \text{Constant times a function property} \\ =5(3^2) & \text{Function raised to an exponent property} \\ =45 & \end{array}

Evaluatelimx4(x35)\lim_{x\to4}(x^3-5).

Example. Evaluate limx5(2x33x+1)\lim_{x\to5}(2x^3-3x+1).

Solution.

limx5(2x33x+1)=limx5(2x3)limx5(3x)+limx5(1)Sum of functions=2limx5(x3)3limx5(x)+limx5(1)Constant times a function=2(53)3(5)+1Function raised to an exponent=236Evaluate \begin{array}{ll} \lim\limits_{x\to5}(2x^3-3x+1)=\lim\limits_{x\to5}(2x^3)-\lim\limits_{x\to5}(3x)+\lim\limits_{x\to5}(1) & \text{Sum of functions} \\ =2\lim\limits_{x\to5}(x^3)-3\lim\limits_{x\to5}(x)+\lim\limits_{x\to5}(1) & \text{Constant times a function} \\ =2(5^3)-3(5)+1 & \text{Function raised to an exponent} \\ =236 & \text{Evaluate} \end{array}

Evaluate the following limit:limx1(x44x3+5)\lim_{x\to-1}(x^4-4x^3+5).

Finding the Limit of a Power or a Root

When a limit includes a power or a root, we need another property to help us evaluate it. The square of the limit of a function equals the limit of the square of the function; the same goes for higher powers. Likewise, the square root of the limit of a function equals the limit of the square root of the function; the same holds true for higher roots.

Example. Evaluate limx2(3x+1)5\lim_{x\to2}(3x+1)^5.

Solution. We will take the limit of the function as xx approaches 2 and raise the result to the 5th power.

limx2(3x+1)5=(limx2(3x+1))5=(3(2)+1)5=75=16,807 \begin{array}{ll} \lim\limits_{x\to2}(3x+1)^5=\left(\lim\limits_{x\to2}(3x+1)\right)^5 & \\ =(3(2)+1)^5 & \\ =7^5 & \\ =16{,}807 & \end{array}

Evaluate the following limit:limx4(10x+36)3\lim_{x\to-4}(10x+36)^3.

Q&A. If we can’t directly apply the properties of a limit, for example in limx2(x2+6x+8x2)\lim_{x\to2}\left(\tfrac{x^2+6x+8}{x-2}\right), can we still determine the limit of the function as xx approaches aa?

Yes. Some functions may be algebraically rearranged so that one can evaluate the limit of a simplified equivalent form of the function.

Finding the Limit of a Quotient

Finding the limit of a function expressed as a quotient can be more complicated. We often need to rewrite the function algebraically before applying the properties of a limit. If the denominator evaluates to 0 when we apply the properties of a limit directly, we must rewrite the quotient in a different form. One approach is to write the quotient in factored form and simplify.

How To: Given the limit of a function in quotient form, use factoring to evaluate it.

  1. Factor the numerator and denominator completely.
  2. Simplify by dividing any factors common to the numerator and denominator.
  3. Evaluate the resulting limit, remembering to use the correct domain.

Example. Evaluate limx2(x26x+8x2)\lim_{x\to2}\left(\tfrac{x^2-6x+8}{x-2}\right).

Solution. Factor where possible, and simplify.

limx2(x26x+8x2)=limx2((x2)(x4)x2)Factor the numerator=limx2(x4)Cancel the common factors=24=2 \begin{array}{ll} \lim\limits_{x\to2}\left(\tfrac{x^2-6x+8}{x-2}\right)=\lim\limits_{x\to2}\left(\tfrac{(x-2)(x-4)}{x-2}\right) & \text{Factor the numerator} \\ =\lim\limits_{x\to2}(x-4) & \text{Cancel the common factors} \\ =2-4=-2 & \end{array}

Analysis. When the limit of a rational function cannot be evaluated directly, factored forms of the numerator and denominator may simplify to a result that can be evaluated. Notice, the function f(x)=x26x+8x2f(x)=\tfrac{x^2-6x+8}{x-2} is equivalent to the function f(x)=x4, x2f(x)=x-4,\ x\ne2. Notice that the limit exists even though the function is not defined at x=2x=2.

Evaluate the following limit:limx7(x211x+287x)\lim_{x\to7}\left(\tfrac{x^2-11x+28}{7-x}\right).

Example. Evaluate limx5(1x15x5)\lim_{x\to5}\left(\tfrac{\tfrac1x-\tfrac15}{x-5}\right).

Solution. Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have the LCD as their denominator.

limx5(1x15x5)=limx5(5x(1x15)5x(x5))Multiply the numerator and denominator by the LCD=limx5(5x(1x)5x(15)5x(x5))Apply the distributive property=limx5(5x5x(x5))Simplify=limx5(1(x5)5x(x5))Factor the numerator=limx5(15x)Cancel the common factor=15(5)Evaluate for x=5=125 \begin{array}{ll} \lim\limits_{x\to5}\left(\tfrac{\tfrac1x-\tfrac15}{x-5}\right)=\lim\limits_{x\to5}\left(\tfrac{5x\left(\tfrac1x-\tfrac15\right)}{5x(x-5)}\right) & \text{Multiply the numerator and denominator by the LCD} \\ =\lim\limits_{x\to5}\left(\tfrac{5x\left(\tfrac1x\right)-5x\left(\tfrac15\right)}{5x(x-5)}\right) & \text{Apply the distributive property} \\ =\lim\limits_{x\to5}\left(\tfrac{5-x}{5x(x-5)}\right) & \text{Simplify} \\ =\lim\limits_{x\to5}\left(\tfrac{-1(x-5)}{5x(x-5)}\right) & \text{Factor the numerator} \\ =\lim\limits_{x\to5}\left(-\tfrac{1}{5x}\right) & \text{Cancel the common factor} \\ =-\tfrac{1}{5(5)} & \text{Evaluate for } x=5 \\ =-\tfrac{1}{25} & \end{array}

Analysis. When determining the limit of a rational function that has terms added or subtracted in either the numerator or denominator, the first step is to find the common denominator of the added or subtracted terms; then, convert both terms to have that denominator, or simplify the rational function by multiplying numerator and denominator by the least common denominator. Then check to see if the resulting numerator and denominator have any common factors.

Evaluatelimx5(15+1x10+2x)\lim_{x\to-5}\left(\tfrac{\tfrac15+\tfrac1x}{10+2x}\right). Enter the answer as a fraction.

How To: Given a limit of a function containing a root, use a conjugate to evaluate.

  1. If the quotient as given is not in indeterminate (00\tfrac00) form, evaluate directly.
  2. Otherwise, rewrite the sum (or difference) of two quotients as a single quotient, using the least common denominator (LCD).
  3. If the numerator includes a root, rationalize the numerator; multiply the numerator and denominator by the conjugate of the numerator. Recall that a±ba\pm\sqrt{b} are conjugates.
  4. Simplify.
  5. Evaluate the resulting limit.

Example. Evaluate limx0(25x5x)\lim_{x\to0}\left(\tfrac{\sqrt{25-x}-5}{x}\right).

Solution.

limx0(25x5x)=limx0(25x5x25x+525x+5)Multiply the numerator and denominator by the conjugate=limx0((25x)25x(25x+5))Multiply: (25x5)(25x+5)=(25x)25=limx0(xx(25x+5))Combine like terms=limx0(125x+5)Simplify xx=1=1250+5Evaluate=15+5=110 \begin{array}{ll} \lim\limits_{x\to0}\left(\tfrac{\sqrt{25-x}-5}{x}\right)=\lim\limits_{x\to0}\left(\tfrac{\sqrt{25-x}-5}{x}\cdot\tfrac{\sqrt{25-x}+5}{\sqrt{25-x}+5}\right) & \text{Multiply the numerator and denominator by the conjugate} \\ =\lim\limits_{x\to0}\left(\tfrac{(25-x)-25}{x(\sqrt{25-x}+5)}\right) & \text{Multiply: }(\sqrt{25-x}-5)(\sqrt{25-x}+5)=(25-x)-25 \\ =\lim\limits_{x\to0}\left(\tfrac{-x}{x(\sqrt{25-x}+5)}\right) & \text{Combine like terms} \\ =\lim\limits_{x\to0}\left(\tfrac{-1}{\sqrt{25-x}+5}\right) & \text{Simplify }\tfrac{-x}{x}=-1 \\ =\tfrac{-1}{\sqrt{25-0}+5} & \text{Evaluate} \\ =\tfrac{-1}{5+5}=-\tfrac{1}{10} & \end{array}

Analysis. When determining a limit of a function with a root as one of two terms where we cannot evaluate directly, think about multiplying the numerator and denominator by the conjugate of the terms.

Evaluate the following limit:limh0(16h4h)\lim_{h\to0}\left(\tfrac{\sqrt{16-h}-4}{h}\right). Enter the answer as a fraction.

Example. Evaluate limx4(4xx2)\lim_{x\to4}\left(\tfrac{4-x}{\sqrt{x}-2}\right).

Solution.

limx4(4xx2)=limx4((2+x)(2x)x2)Factor=limx4((2+x)(2x)(2x))Factor 1 out of the denominator; simplify=limx4(2+x)Evaluate=(2+4)=4 \begin{array}{ll} \lim\limits_{x\to4}\left(\tfrac{4-x}{\sqrt{x}-2}\right)=\lim\limits_{x\to4}\left(\tfrac{(2+\sqrt{x})(2-\sqrt{x})}{\sqrt{x}-2}\right) & \text{Factor} \\ =\lim\limits_{x\to4}\left(\tfrac{(2+\sqrt{x})(2-\sqrt{x})}{-(2-\sqrt{x})}\right) & \text{Factor }-1\text{ out of the denominator; simplify} \\ =\lim\limits_{x\to4}-(2+\sqrt{x}) & \text{Evaluate} \\ =-(2+\sqrt4) & \\ =-4 & \end{array}

Analysis. Multiplying by a conjugate would expand the numerator; look instead for factors in the numerator. Four is a perfect square so that the numerator is in the form a2b2a^2-b^2 and may be factored as (a+b)(ab)(a+b)(a-b).

Evaluate the following limit:limx3(x3x3)\lim_{x\to3}\left(\tfrac{x-3}{\sqrt{x}-\sqrt{3}}\right).

How To: Given a quotient with absolute values, evaluate its limit.

  1. Try factoring or finding the LCD.
  2. If the limit cannot be found, choose several values close to and on either side of the input where the function is undefined.
  3. Use the numeric evidence to estimate the limits on both sides.

Example. Evaluate limx7x7x7\lim_{x\to7}\tfrac{|x-7|}{x-7}.

Solution. The function is undefined at x=7x=7, so we will try values close to 7 from the left and the right.

Left-hand limit: 6.976.97=6.9976.997=6.99976.9997=1\tfrac{|6.9-7|}{6.9-7}=\tfrac{|6.99-7|}{6.99-7}=\tfrac{|6.999-7|}{6.999-7}=-1

Right-hand limit: 7.177.17=7.0177.017=7.00177.0017=1\tfrac{|7.1-7|}{7.1-7}=\tfrac{|7.01-7|}{7.01-7}=\tfrac{|7.001-7|}{7.001-7}=1

Since the left- and right-hand limits are not equal, there is no limit.

Evaluatelimx6+6xx6\lim_{x\to6^+}\tfrac{6-x}{|x-6|}(the right-hand limit asxxapproaches66).

Media. Access the following online resource for additional instruction and practice with properties of limits.

Key concepts

  • The properties of limits can be used to perform operations on the limits of functions rather than the functions themselves.
  • The limit of a polynomial function can be found by finding the sum of the limits of the individual terms.
  • The limit of a function that has been raised to a power equals the same power of the limit of the function. Another method is direct substitution.
  • The limit of the root of a function equals the corresponding root of the limit of the function.
  • One way to find the limit of a function expressed as a quotient is to write the quotient in factored form and simplify.
  • Another method of finding the limit of a complex fraction is to find the LCD.
  • A limit containing a function containing a root may be evaluated using a conjugate.
  • The limits of some functions expressed as quotients can be found by factoring.
  • One way to evaluate the limit of a quotient containing absolute values is by using numeric evidence. Setting it up piecewise can also be useful.

Practice

Find the limit of a sum, a difference, and a product

For the limitlimx3(2x+5)=11\lim_{x\to3}(2x+5)=11, worked out using the Sum of functions and Constant times a function properties, identifyaa.

For the limitlimx3(2x+5)=11\lim_{x\to3}(2x+5)=11, identifyf(x)f(x).

For the limitlimx3(2x+5)=11\lim_{x\to3}(2x+5)=11, identifyLL, the value of the limit.

What does it mean to say the limit off(x)f(x), asxxapproachescc, is undefined?

Find the limit of a polynomial

Give an example of a type of functionffwhose limit, asxxapproachesaa, isf(a)f(a).

Letf(x)={2x2+2x+1,x0x3,x>0f(x)=\begin{cases} 2x^2+2x+1, & x\le0 \\ x-3, & x>0 \end{cases}. Findlimx0+f(x)\lim_{x\to0^+}f(x).

Forf(x)={2x2+2x+1,x0x3,x>0f(x)=\begin{cases} 2x^2+2x+1, & x\le0 \\ x-3, & x>0 \end{cases}, what islimx0f(x)\lim_{x\to0}f(x)?

Find the limit of a power or a root

Givenlimxcf(x)=3\lim_{x\to c}f(x)=3andlimxcg(x)=5\lim_{x\to c}g(x)=5, evaluatelimxc[2f(x)+g(x)]\lim_{x\to c}\left[2f(x)+\sqrt{g(x)}\right].

For the limitlimx2(3x+1)5=16,807\lim_{x\to2}(3x+1)^5=16{,}807, identifyaa.

For the limitlimx2(3x+1)5=16,807\lim_{x\to2}(3x+1)^5=16{,}807, identifyf(x)f(x), the function whose limit is being taken.

For the limitlimx2(3x+1)5=16,807\lim_{x\to2}(3x+1)^5=16{,}807, identifyLL, the value of the limit.

Find the limit of a quotient

Evaluate:limx3(x29x3)\lim_{x\to3}\left(\tfrac{x^2-9}{x-3}\right).

Evaluate:limx32(6x217x+122x3)\lim_{x\to\frac32}\left(\tfrac{6x^2-17x+12}{2x-3}\right). Enter the answer as a fraction.

Evaluate:limx3(x2+2x3x3)\lim_{x\to3}\left(\tfrac{x^2+2x-3}{x-3}\right).

Forf(x)=x2+4x100f(x)=x^2+4x-100, find the average rate of changef(x+h)f(x)h\tfrac{f(x+h)-f(x)}{h}.

Forf(x)=1xf(x)=\tfrac1x, find the average rate of changef(x+h)f(x)h\tfrac{f(x+h)-f(x)}{h}.

Forf(x)=xf(x)=\sqrt{x}, find the average rate of changef(x+h)f(x)h\tfrac{f(x+h)-f(x)}{h}.

Which equation could be represented by the graph below?

For the following exercise, refer to the graph below.

What is the left-hand limit of the function graphed above asxxapproaches 0?


This section is adapted from Precalculus 2e, Section 12.2: Finding Limits: Properties of Limits by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated Figures 1 and 2 (the hole in f(x)=x26x7x7f(x)=\tfrac{x^2-6x-7}{x-7} at x=7x=7 and the continuous line g(x)=x+1g(x)=x+1) as accessible spec-first SVGs, each an exact line with slope 11 and intercept 11, checked against the rendered PDF, page 1188 (true PDF index 1198). The pinned CNXML’s <solution> for Example 6 (“Evaluating the Limit of a Quotient by Finding the LCD”) is a single JPEG image of the worked steps rather than transcribable text or MathML, so its seven-row step table was transcribed instead from the rendered PDF, page 1192 (true PDF index 1202), and independently re-derived; the numeric result, 125-\tfrac{1}{25}, matches both the image and the independent derivation. Omitted the “Media” callout’s one external video link, matching house precedent elsewhere in this book. Every “evaluate the limit” fill-in declares an answerForm (decimal for an integer or terminating-decimal result, fraction for a fraction result with “enter the answer as a fraction” stated in the question, exact-radical for a bare radical result) because the grading engine evaluates a retyped \lim expression rather than merely comparing it to the printed span — confirmed by replaying each printed limit span against its own key — and a limit whose value is “does not exist” is a multiplechoice throughout, since MathLive cannot type words and the engine cannot grade text. One-sided limits (the Try It following Example 9, and the Practice item reading the section’s third recreated graph) are asked in words rather than with the unparseable limxa+\lim_{x\to a^+} notation, per this book’s established convention. The two Verbal exercises with solutions (on what an undefined limit means, and on which function type satisfies limxaf(x)=f(a)\lim_{x\to a}f(x)=f(a)) were adapted into multiple-choice questions for the “sum, difference, and product” and “polynomial” objective groups, since the Algebraic exercise set turned out, on inspection, to contain no plain sum/difference/product or pure-polynomial limit beyond what the worked examples already cover — every numbered Algebraic exercise with a solution is a rational-function quotient. The “sum, difference, and product” and “power or root” groups are filled out with an “identify aa, f(x)f(x), and LL” notation-reading triple of fill-ins apiece, drawn from Examples 1 and 4 of this section — the fallback this book’s authoring playbook names for a thin objective group with no dedicated source exercise. The “polynomial” objective’s group repurposes Algebraic exercises 37 and 39 (the piecewise function’s right-hand and two-sided limits at x=0x=0), whose branches are themselves polynomials evaluated by direct substitution. Thirteen selected end-of-section Algebraic, Graphical, and Verbal exercises with a printed official answer were adapted into interactive Practice components across the four objective groups; the two Graphical exercises without a solution in the Answer Key (exercise 54, “find an equation represented by” the removable-discontinuity graph at (2,3)(2,3), and exercise 56, the same graph’s right-hand limit) were independently derivable but omitted per this book’s exercise-selection rule, which uses only items whose answer is visibly present in the source. The Practice item for f(x)=xf(x)=\sqrt{x}’s average rate of change is a multiple choice, since every declared answerForm fails open on the retyped difference-quotient span — confirmed by replay — and its correct option, 1x+h+x\tfrac{1}{\sqrt{x+h}+\sqrt{x}}, corrects a sign error in the pinned CNXML’s own printed key, 1x+h+x\tfrac{-1}{\sqrt{x+h}+\sqrt{x}} (the difference quotient of x\sqrt{x} rationalizes to hh(x+h+x)\tfrac{h}{h(\sqrt{x+h}+\sqrt{x})}, which is positive); the distractor options include the source’s own printed (incorrect) sign. The Graphical exercise “find an equation that could be represented by” the graph with a hole at (3,1)(-3,-1) is a multiple choice rather than a fill-in, because the grading engine proves a rational function equal to its reduced form — x+2x+2, and even a quotient whose removable hole sits at a different xx, grade equal to the keyed x2+5x+6x+3\tfrac{x^2+5x+6}{x+3} — so no fill-in could distinguish the answer from a function without the hole.