math.wiki
Calculus › Limits › What a limit is: graphs and tables
Open on the map
Precalculus

What a limit is: graphs and tables

A formula can refuse an input, or have its value there changed, and still have outputs near that input crowding around one height. How do you name that height, and how do you read it off a graph or a table?

Where you are 11 in Limits

When you reach for it

To know what a function does at an input, you evaluate it. Often the question is what it does near the input. The formula \displaystyle \frac{x^2 - x - 6}{x - 3} refuses x = 3, yet its outputs near 3 crowd around one height. In Rates at an instant from shrinking intervals a speed at an instant was estimated by average speeds over shrinking intervals, and the estimates closed in on one number. A graph settling onto a horizontal asymptote and the sum of an infinite geometric series are numbers that outputs approach without reaching them in one step. The limit is the single idea under all of these: the height a function heads toward as its input heads toward a target.

The lesson takes two days. Part 1 is the idea of a limit, one-sided limits, and how a limit differs from the value of the function; Examples 1 to 4 go with it. Part 2 is limits that do not exist, limits as x \to \infty, and the infinite sum as a limit; Examples 5 to 7 go with it.

The procedure

Part 1: the idea of a limit

Definition. Let f be defined at every input near a on both sides, except perhaps at a itself. Suppose that for every input x \ne a close enough to a, the output f(x) is as close as you like to one number L. Then L is the limit of f(x) as x approaches a, written

\lim_{x \to a} f(x) = L

Three things in the definition matter. The input approaches a from both sides. The input never equals a, so the value f(a), if there is one, plays no part. And “as close as you like” means any closeness at all can be demanded, and it is met once x is near enough to a. Making that sentence exact, with a named tolerance and a window of inputs, is the epsilon delta definition met in calculus.

Reading a limit from a graph. Put a finger on the graph a little to the left of x = a and slide it toward a along the curve. Do the same from the right. If both fingers head toward the same point (a, L), the limit is L. Whatever sits directly above a (a closed dot, an open circle, a dot at another height, or nothing) is ignored, because the definition never looks at x = a.

One-sided limits. Often the two sides need to be watched separately. Write

\lim_{x \to a^-} f(x) = L

when the outputs approach L as x approaches a through inputs less than a, from the left; \lim_{x \to a^+} f(x) is the same from the right. This is the notation of Continuous graphs, jumps and holes and Vertical asymptotes. The two-sided limit asks both sides to close in on one number, so the limit \lim_{x \to a} f(x) exists exactly when both one-sided limits exist and are equal, and then it is their common value. This is One sided limits and when a limit exists.

Limit versus value. Figure 1 shows the same curve three ways at x = a.

a x y L A a x y L B a x y L C
Figure 1In all three pictures \lim_{x \to a} f(x) = L. The value f(a) is L in A, undefined in B, and a different number in C.

In A the graph passes through (a, L), so the limit and the value are both L. In B the point is missing, a hole: the limit is L but f(a) is undefined. Holes appear in rational functions where a factor cancels, as in Graphing rational functions. In C the point has been moved: f(a) is another number, and the limit is still L. The limit and the value answer different questions. When they agree, as in A, the function is continuous at a; that is the definition of Continuity, met informally in Continuous graphs, jumps and holes.

Estimating a limit from a table. With a formula and no graph, evaluate f at inputs closing in on a from both sides, such as a - 0.1, a - 0.01, a - 0.001 and a + 0.001, a + 0.01, a + 0.1. If the outputs on both sides settle on the same digits, those digits estimate the limit. A table is evidence, not proof: it checks a few inputs, and the definition speaks of all of them near a. Example 6 shows a table that looks convincing and is wrong. When the formula can be simplified, for instance by canceling a factor, the algebra confirms the estimate; that is Limits by algebra: factoring and rationalizing.

Part 2: limits that do not exist, and limits at infinity

A limit fails to exist when the outputs near a do not close in on one number. There are three ways this happens.

  • A jump. Both one-sided limits exist but are different numbers, so no single L serves both sides. See Discontinuities and how to remove them.
  • Unbounded growth. Near a vertical asymptote the outputs pass every number. Writing \lim_{x \to a^+} f(x) = \infty records how the limit fails: the outputs grow without bound through positive values. The symbol \infty is not a number, so the limit still does not exist. If both sides go to \infty, write \lim_{x \to a} f(x) = \infty. This is Infinite limits and vertical asymptotes.
  • Endless oscillation. However close x comes to a, the outputs keep swinging between heights that stay apart, as \displaystyle \sin\left(\frac{1}{x}\right) does near 0.

Limits at infinity. \lim_{x \to \infty} f(x) = L means that the outputs are as close as you like to L for every large enough x. On a graph the curve settles onto the horizontal line y = L far to the right, a horizontal asymptote. The limit as x \to -\infty is the same far to the left. A table uses inputs such as 10, 100, 1000.

An infinite sum as a limit. A sequence is a function of n = 1, 2, 3, \ldots, so the same idea applies as n \to \infty. The partial sums S_n of a series form a sequence, and the sum of the series is

S = \lim_{n \to \infty} S_n

when that limit exists. This is the definition of the sum in Infinite geometric series, written in limit notation. For a geometric series with |r| < 1, \displaystyle S_n = \frac{a_1(1 - r^n)}{1 - r} and r^n approaches 0, so \displaystyle S = \frac{a_1}{1 - r}. When |r| \ge 1 and a_1 \ne 0 the partial sums have no limit and the series has no sum.

Worked examples

Example 1

Reading limits and values from a graph

Figure 2 is the graph of a function f. For each of a = -1, a = 1 and a = 4, find \lim_{x \to a} f(x) and f(a).

−4 2 4 6 −2 4 x y
(-1, 1)
(1, 3)
(1, -1)
(4, 1)
Figure 2The graph of f in Example 1.
Solution

At x = -1 the graph is one unbroken stroke. Sliding along it from either side reaches the closed dot (-1, 1), so \lim_{x \to -1} f(x) = 1, and the dot is on the graph, so f(-1) = 1. Limit and value agree.

At x = 1, from the left the line heads to the open circle (1, 3), so \lim_{x \to 1^-} f(x) = 3. From the right the curve also starts at that open circle, so \lim_{x \to 1^+} f(x) = 3. The one-sided limits agree, so \lim_{x \to 1} f(x) = 3. The closed dot below says f(1) = -1. This is picture C: a moved point.

At x = 4 the curve heads into the open circle (4, 1) from both sides, so \lim_{x \to 4} f(x) = 1. No dot sits anywhere else above 4, so f(4) is undefined. This is picture B: a hole.

Example 2

Inocencia’s table at a refused input

Inocencia wants \lim_{x \to 3} f(x) for \displaystyle f(x) = \frac{x^2 - x - 6}{x - 3}. Estimate it from a table, then confirm it.

Solution

At x = 3 both top and bottom are 0, and \displaystyle \frac{0}{0} is not a number, so f(3) is undefined. That says nothing yet about the limit. For instance, \displaystyle f(2.9) = \frac{8.41 - 2.9 - 6}{-0.1} = \frac{-0.49}{-0.1} = 4.9, and the rest of the table is found the same way.

x (left)f(x)x (right)f(x)
2.94.93.15.1
2.994.993.015.01
2.9994.9993.0015.001

From both sides the outputs settle toward 5. To confirm, factor the top: x^2 - x - 6 = (x - 3)(x + 2). For x \ne 3 the factor x - 3 is not zero, so it cancels and f(x) = x + 2. Then the distance from f(x) to 5 is exactly |x - 3|, which is as small as you like once x is close to 3. So \lim_{x \to 3} f(x) = 5, while f(3) stays undefined. The graph is the line y = x + 2 with a hole at (3, 5).

Example 3

Primitivo’s table with no algebra shortcut

Primitivo estimates \lim_{x \to 0} g(x) for \displaystyle g(x) = \frac{3^x - 1}{x}, which is \displaystyle \frac{0}{0} at x = 0.

Solution

No factor cancels here, so the table is the main tool. A calculator gives, to five decimal places:

x (left)g(x)x (right)g(x)
-0.11.040420.11.16123
-0.011.092600.011.10467
-0.0011.098010.0011.09922

The left outputs rise and the right outputs fall, and the gap between them shrinks about tenfold with each row: 0.12081, then 0.01207, then 0.00121. The outputs from both sides are closing in on a number near 1.099, which is the estimate. The exact value is \ln 3 \approx 1.098612, a fact calculus proves. The quotient \displaystyle \frac{3^x - 1}{x} is also the average rate of change of 3^x between 0 and x, so this limit is the rate at which 3^x grows at the instant x = 0.

Example 4

One-sided limits of a piecewise function

Let

p(x) = \begin{cases} x^2 + 1, & x < 2 \\ 1, & x = 2 \\ 7 - x, & x > 2 \end{cases}

Find both one-sided limits at 2, the limit, and p(2).

Solution

Inputs just left of 2 use the rule x^2 + 1, a polynomial, whose outputs near 2 are near 2^2 + 1 = 5. So \lim_{x \to 2^-} p(x) = 5. Inputs just right of 2 use 7 - x, near 7 - 2 = 5, so \lim_{x \to 2^+} p(x) = 5. The one-sided limits agree, so \lim_{x \to 2} p(x) = 5. The middle rule gives p(2) = 1. The limit is 5 and the value is 1: a moved point. Had the right rule been 8 - x, the right-hand limit would be 6, the two sides would disagree, and there would be no limit at all, which is where Part 2 begins.

Example 5

A jump, an asymptote and the far right

Figure 3 is the graph of h. Describe h near x = -2 and near x = 2, and find \lim_{x \to \infty} h(x).

−4 −2 8 −2 6 x y
x = 2
y = 1
(-2, 1)
(-2, 3)
Figure 3The graph of h in Example 5. For x > 2 its formula is \displaystyle h(x) = 1 - \frac{4}{x - 2}.
Solution

Near x = -2, from the left the line rises to the open circle (-2, 1), so \lim_{x \to -2^-} h(x) = 1. From the right the curve starts at the closed dot (-2, 3), so \lim_{x \to -2^+} h(x) = 3. The two sides disagree, so \lim_{x \to -2} h(x) does not exist: a jump. Yet h(-2) = 3 is defined. In Example 2 the value was missing and the limit existed; here the value exists and the limit does not.

Near x = 2, from the left the curve climbs along the dashed line past every height, so \lim_{x \to 2^-} h(x) = \infty. From the right it comes up from below every height, so \lim_{x \to 2^+} h(x) = -\infty. The limit at 2 does not exist, and since the sides go opposite ways not even \lim_{x \to 2} h(x) = \infty can be written. Also h(2) is undefined.

Far to the right the curve rises toward the dashed line y = 1 from below, so \lim_{x \to \infty} h(x) = 1. The formula agrees: \displaystyle h(102) = 1 - \frac{4}{100} = 0.96 and \displaystyle h(1002) = 1 - \frac{4}{1000} = 0.996, and the gap \displaystyle \frac{4}{x - 2} shrinks toward 0.

Example 6

An oscillation that fools a table

Inocencia evaluates \displaystyle s(x) = \sin\left(\frac{1}{x}\right) (radians) at \displaystyle x = \frac{1}{10\pi}, \frac{1}{100\pi}, \frac{1}{1000\pi} and at their negatives, and gets 0 every time. Is \lim_{x \to 0} s(x) = 0?

Solution

Her outputs are correct: at \displaystyle x = \frac{1}{10\pi}, \displaystyle \frac{1}{x} = 10\pi and \sin(10\pi) = 0, and the same holds for 100\pi and 1000\pi. But other inputs close in on 0 too. Take \displaystyle x = \frac{2}{41\pi} \approx 0.0155, \displaystyle \frac{2}{401\pi} \approx 0.00159, \displaystyle \frac{2}{4001\pi} \approx 0.000159. Then \displaystyle \frac{1}{x} = \frac{41\pi}{2} = 20\pi + \frac{\pi}{2}, and so on, so every one of these outputs is \displaystyle \sin\left(\frac{\pi}{2}\right) = 1. Inputs \displaystyle x = \frac{2}{43\pi}, \frac{2}{403\pi}, \ldots give \displaystyle \sin\left(\frac{3\pi}{2}\right) = -1.

In fact, for x between 0 and any small \delta > 0, the number \displaystyle \frac{1}{x} runs through every value above \displaystyle \frac{1}{\delta}, which includes whole turns of the unit circle, so s takes both 1 and -1 there. No number L is within \displaystyle \frac{1}{2} of both 1 and -1, so the outputs cannot stay close to any single L. The limit does not exist, and neither one-sided limit does (Figure 4). Inocencia’s inputs all happened to land on zeros of sine.

−0.4 −0.2 0.2 0.4 x y
y = 1
y = -1
Figure 4The graph of \displaystyle s(x) = \sin\left(\frac{1}{x}\right). Near 0 it touches 1 and -1 infinitely often.
Example 7

The sum of a series as a limit

Gumersinda adds \displaystyle 5 + 2 + \frac{4}{5} + \frac{8}{25} + \cdots. Tabulate the partial sums and find \lim_{n \to \infty} S_n.

Solution

Each term is \displaystyle \frac{2}{5} of the one before, so the series is geometric with a_1 = 5 and \displaystyle r = \frac{2}{5}. The finite sum formula gives

S_n = \frac{5\left(1 - \left(\frac{2}{5}\right)^n\right)}{\frac{3}{5}} = \frac{25}{3}\left(1 - \left(\frac{2}{5}\right)^n\right)
n12345
S_n577.88.128.248

The gap \displaystyle \frac{25}{3} - S_n = \frac{25}{3}\left(\frac{2}{5}\right)^n is positive and shrinks by the factor \displaystyle \frac{2}{5} at every step, so it becomes as small as you like. Hence \displaystyle \lim_{n \to \infty} S_n = \frac{25}{3}, and the sum of the series is \displaystyle \frac{25}{3} \approx 8.333. By contrast, the partial sums of 4 - 4 + 4 - 4 + \cdots are 4, 0, 4, 0, \ldots; they jump between two heights for ever, so they have no limit and that series has no sum.

Boundaries

  • The two-sided limit needs f defined on both sides near a. Where inputs exist on one side only, as for \displaystyle \sqrt{x} at 0, only the one-sided limit is asked: \displaystyle \lim_{x \to 0^+} \sqrt{x} = 0.
  • A limit, when it exists, is one number. “\lim_{x \to a} f(x) = \infty” is not a limit; it says how the limit fails.
  • The limit at a says nothing about f(a), and f(a) says nothing about the limit.
  • A table never proves a limit (Example 6). A computer graph can hide a hole, since one missing point has no width on the screen, and cannot draw an oscillation like Figure 4 faithfully near 0.
  • A graph may cross its horizontal asymptote and still have \lim_{x \to \infty} f(x) = L; only the far-out behavior counts.

Where this goes next

In calculus limits are found exactly rather than estimated. The limit laws say limits respect sums and products, Limits by algebra: factoring and rationalizing removes the \displaystyle \frac{0}{0} of Example 2, and The squeeze theorem and the limit of sine x over x settles a limit no algebra reaches. Continuity is defined as “the limit equals the value”. The shrinking intervals of Rates at an instant from shrinking intervals become the derivative, a limit of average rates, and areas become the limit of sums of rectangles. Far-out and blow-up behavior return as Limits at infinity and horizontal asymptotes and Infinite limits and vertical asymptotes. Finally, The epsilon delta definition of a limit makes “as close as you like” exact.

Exercises

Part 1: reading the graph of k

−2 2 4 6 8 −4 −2 6 x y
x = 5
y = 2
Figure 5The graph of k for Exercises 1 to 3 and 10 to 12. For x > 5 its formula is \displaystyle k(x) = 2 + \frac{3}{x - 5}.
  1. Find \lim_{x \to -3} k(x) and k(-3).
  2. Find \lim_{x \to -1^-} k(x), \lim_{x \to -1^+} k(x), \lim_{x \to -1} k(x) and k(-1).
  3. Find \lim_{x \to 1} k(x) and k(1).

Part 1: tables and formulas

  1. For \displaystyle f(x) = \frac{x^2 + 2x - 8}{x - 2}, find f(1.9), f(1.99), f(2.01) and f(2.1), estimate \lim_{x \to 2} f(x), and confirm it by factoring.
  2. Find \displaystyle \lim_{x \to -5} \frac{x^2 - 25}{x + 5}, using the inputs -5.1 and -4.9 to check.
  3. Estimate \displaystyle \lim_{x \to 0} \frac{\sqrt{x + 9} - 3}{x} from its values at x = \pm 0.01 and x = \pm 0.001, to five decimal places.
  4. Estimate \displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x^2} (x in radians) from its values at x = 0.1 and x = 0.01, to six decimal places. Why do the negative inputs give nothing new?
  5. Let q(x) = 2x + 1 for x < 3, q(3) = 4, and q(x) = 10 - x for x > 3. Find both one-sided limits at 3, the limit, and q(3).
  6. Let \displaystyle r(x) = \frac{x^2 - 1}{x - 1} for x \ne 1 and r(1) = 5. Find \lim_{x \to 1} r(x) and r(1).

Part 2: reading the graph of k

  1. Find \lim_{x \to 3^-} k(x), \lim_{x \to 3^+} k(x), \lim_{x \to 3} k(x) and k(3).
  2. Describe k(x) as x \to 5^- and as x \to 5^+. Does \lim_{x \to 5} k(x) exist?
  3. Find \lim_{x \to \infty} k(x), and check with the formula at x = 105.

Part 2: formulas and tables

  1. Let \displaystyle j(x) = \frac{|x - 4|}{x - 4}. Find \lim_{x \to 4^-} j(x), \lim_{x \to 4^+} j(x) and \lim_{x \to 4} j(x).
  2. Describe \displaystyle \frac{1}{(x + 1)^2} as x \to -1, using the inputs -1.01 and -0.99.
  3. Evaluate \displaystyle \frac{5x - 2}{x + 1} at x = 10, 100 and 1000 (five decimal places) and estimate its limit as x \to \infty.
  4. Find \lim_{x \to -\infty} 2^x, and describe 2^x as x \to \infty.
  5. Show that \displaystyle \lim_{x \to 0} \cos\left(\frac{1}{x}\right) does not exist by evaluating at \displaystyle x = \frac{1}{2m\pi} and at \displaystyle x = \frac{1}{(2m + 1)\pi} for positive integers m.

Part 2: infinite sums

  1. For \displaystyle 9 + 6 + 4 + \frac{8}{3} + \cdots, find S_1 to S_4 and \lim_{n \to \infty} S_n.
  2. The repeating decimal 0.272727\ldots is the limit of the partial sums of 0.27 + 0.0027 + 0.000027 + \cdots. Find it as a fraction.
  3. Find the partial sums of 2 - 2 + 2 - 2 + \cdots. Do they have a limit?

Reasoning

  1. Primitivo says “f(3) = 7, so \lim_{x \to 3} f(x) = 7.” Inocencia says “\lim_{x \to 3} f(x) = 7, so f(3) = 7.” Describe a graph that shows each claim is false.
  2. Explain why a function cannot have \lim_{x \to a} f(x) = 3 and \lim_{x \to a} f(x) = 4 at once.
  3. Gumersinda’s table of some f at x = 0.1, 0.01, 0.001 shows 2.1, 2.01, 2.001. Give two reasons this alone does not show \lim_{x \to 0} f(x) = 2.
  4. Use partial sums to explain why a_1 + a_1 + a_1 + \cdots (ratio r = 1, a_1 \ne 0) has no sum.
Answers
  1. \lim_{x \to -3} k(x) = 1 and k(-3) = 1. The line is unbroken at -3 and passes through (-3, 1).
  2. 3, 3, 3, and k(-1) = -1. Both pieces head to the open circle (-1, 3); the closed dot is at (-1, -1), a moved point.
  3. \lim_{x \to 1} k(x) = 2 and k(1) = 2. The falling line from (-1, 3) to (3, 1) drops \displaystyle \frac{1}{2} per unit, so it is at height 2 above 1.
  4. 5.9, 5.99, 6.01, 6.1; the limit is 6. For x \ne 2, \displaystyle \frac{(x - 2)(x + 4)}{x - 2} = x + 4, which is near 6 when x is near 2.
  5. -10. The outputs are -10.1 and -9.9; for x \ne -5 the quotient is \displaystyle \frac{(x - 5)(x + 5)}{x + 5} = x - 5, near -10.
  6. About 0.1667. Values: 0.16671 at -0.01, 0.16667 at -0.001, 0.16666 at 0.001, 0.16662 at 0.01. Multiplying top and bottom by \displaystyle \sqrt{x + 9} + 3 gives \displaystyle \frac{1}{\sqrt{x + 9} + 3}, so the limit is exactly \displaystyle \frac{1}{6}.
  7. About 0.5. Values: 0.499583 at 0.1 and 0.499996 at 0.01. Since \cos(-x) = \cos x and (-x)^2 = x^2, each negative input gives the same output as its positive partner.
  8. Both one-sided limits are 7 (2 \cdot 3 + 1 = 7 and 10 - 3 = 7), so the limit is 7, while q(3) = 4.
  9. \lim_{x \to 1} r(x) = 2, since \displaystyle \frac{(x - 1)(x + 1)}{x - 1} = x + 1 for x \ne 1; and r(1) = 5.
  10. 1, -2, does not exist (a jump), and k(3) = -2.
  11. k(x) \to -\infty as x \to 5^- and k(x) \to \infty as x \to 5^+. The limit does not exist.
  12. 2. The curve settles onto y = 2, and \displaystyle k(105) = 2 + \frac{3}{100} = 2.03.
  13. -1, 1, does not exist. Left of 4 the top is -(x - 4), so j(x) = -1; right of 4, j(x) = 1.
  14. Both inputs give \displaystyle \frac{1}{0.0001} = 10{,}000. The outputs grow without bound from both sides, so \displaystyle \lim_{x \to -1} \frac{1}{(x + 1)^2} = \infty: no limit as a number.
  15. 4.36364, 4.93069, 4.99301; the limit is 5. Indeed \displaystyle \frac{5x - 2}{x + 1} = 5 - \frac{7}{x + 1}.
  16. \lim_{x \to -\infty} 2^x = 0, since for negative x, \displaystyle 2^x = \frac{1}{2^{|x|}} and 2^{|x|} grows without bound. As x \to \infty, 2^x grows without bound, so there is no limit: \lim_{x \to \infty} 2^x = \infty.
  17. At \displaystyle x = \frac{1}{2m\pi} the output is \cos(2m\pi) = 1; at \displaystyle x = \frac{1}{(2m + 1)\pi} it is \cos((2m + 1)\pi) = -1. Both kinds of input close in on 0, so the outputs cannot stay close to one number.
  18. S_1 = 9, S_2 = 15, S_3 = 19, \displaystyle S_4 = \frac{65}{3}. With \displaystyle r = \frac{2}{3}, \displaystyle \lim_{n \to \infty} S_n = \frac{9}{1 - \frac{2}{3}} = 27.
  19. \displaystyle \frac{3}{11}. Here a_1 = 0.27 and r = 0.01, so the limit is \displaystyle \frac{0.27}{0.99} = \frac{27}{99} = \frac{3}{11}.
  20. 2, 0, 2, 0, \ldots They jump between two heights for ever and approach no single number, so there is no limit and the series has no sum.
  21. Primitivo: a graph with a closed dot at (3, 7) and a jump there, for example heading to 7 from the left and to 1 from the right, has f(3) = 7 but no limit. Inocencia: a graph through an open circle at (3, 7) with a closed dot at (3, 0), or no dot at all, has limit 7 but f(3) \ne 7.
  22. Near a, the outputs would have to be within \displaystyle \frac{1}{4} of 3 and within \displaystyle \frac{1}{4} of 4 at the same time. No number is, since 3 and 4 are 1 apart.
  23. The inputs come only from the right, so the left side is unknown; and three inputs are not all inputs near 0 (Example 6 shows what can hide between them).
  24. S_n = n a_1, which grows past every number in size as n grows, so the partial sums have no limit and the series has no sum.
unattributed · CC-BY-SA-4.0