Unit 10, lesson 110

Approaching a Limit Numerically and Graphically

How to estimate the limit of a function at a point from tables and graphs, why the value at the point does not matter, how the two sides can disagree, and why evidence is not proof.

  • Theory part 219
  • Practice part 220
  • Limits

Type the formula

f(x)=x2−9x−3f(x) = \frac{x^2 - 9}{x - 3}

into a graphing tool and you see what looks like a straight line. Put in x=3x = 3 and the formula breaks down into 0/00/0. Both observations are right. Near 33 the outputs crowd around 66, and at 33 there is no output at all. A limit describes exactly this difference between near a point and at a point.

Key terms

Nearby inputs

Definition For a number aa and a radius δ>0\delta > 0, the inputs within δ\delta of aa other than aa itself are the numbers xx with

0<∣x−a∣<δ,0 < \lvert x - a \rvert < \delta,

which form the open interval from a−δa - \delta to a+δa + \delta with aa removed. Let ff be defined at all of these inputs for some δ\delta. We say that f(x)f(x) approaches the number LL as xx approaches aa, and write

lim⁡x→af(x)=L,\lim_{x \to a} f(x) = L,

if the outputs can be made as close to LL as we like by taking xx close enough to aa with x≠ax \ne a. Precisely, for every ε>0\varepsilon > 0 there is a δ>0\delta > 0 such that every input with 0<∣x−a∣<δ0 < \lvert x - a \rvert < \delta gives ∣f(x)−L∣<ε\lvert f(x) - L \rvert < \varepsilon. So the limit depends only on the values of ff at nearby inputs, never on f(a)f(a) and never on values far from aa.

In plain words To find a limit at aa, look only at inputs close to aa on both sides and ignore aa itself. Then ask which single number the outputs crowd around.

Example The function from the opening is undefined at 33 and defined everywhere else. Since x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3), it equals x+3x + 3 for every x≠3x \ne 3.

xx2.92.92.992.992.9992.9993.0013.0013.013.013.13.1
f(x)f(x)5.95.95.995.995.9995.9996.0016.0016.016.016.16.1

In fact ∣f(x)−6∣=∣x−3∣\lvert f(x) - 6 \rvert = \lvert x - 3 \rvert for every x≠3x \ne 3, so taking δ=ε\delta = \varepsilon works for every tolerance. The limit is 66.

Function value versus limiting value

Definition The function value f(a)f(a) is the output at the single input aa, when aa is in the domain. The limiting value is the limit of f(x)f(x) as xx approaches aa, when that limit exists. Either one can exist without the other, and when both exist they can differ. When both exist and are equal, the graph has no break at aa, and calculus calls ff continuous at aa.

In plain words The value f(a)f(a) is where the function is at aa. The limit is where it is heading as xx closes in on aa. Often they agree, but nothing forces them to.

Example All three functions below agree with x2−1x^2 - 1 at every input except possibly 22, so all three have limit 33 at 22.

FunctionValue at 22Limit at 22
x2−1x^2 - 1 for every xx3333
x2−1x^2 - 1 for x≠2x \ne 2, undefined at 22undefined33
x2−1x^2 - 1 for x≠2x \ne 2, and 77 at x=2x = 27733

The three graphs are the full parabola, the parabola with a hole at (2,3)(2, 3), and the parabola with that point moved up to (2,7)(2, 7).

Left and right approach

Definition The left-hand limit

lim⁡x→a−f(x)=L\lim_{x \to a^-} f(x) = L

means that f(x)f(x) approaches LL as xx approaches aa through inputs x<ax < a only. The right-hand limit, written with x→a+x \to a^+, uses inputs x>ax > a only. Each one needs ff to be defined on an open interval just to that side of aa. The two-sided limit exists and equals LL exactly when both one-sided limits exist and both equal LL.

In plain words Walk toward aa from the left and see where the outputs head, then do the same from the right. There is a two-sided limit only when both trips end at the same height.

Example Take the piecewise function

f(x)={x+1,x<1,x2+3,x≥1.f(x) = \begin{cases} x + 1, & x < 1, \\ x^2 + 3, & x \ge 1. \end{cases}
From the left xx0.90.90.990.990.9990.999
f(x)f(x)1.91.91.991.991.9991.999
From the right xx1.11.11.011.011.0011.001
f(x)f(x)4.214.214.02014.02014.0020014.002001

The left-hand limit is 22 and the right-hand limit is 44. They differ, so the two-sided limit at 11 does not exist, even though f(1)=4f(1) = 4 is defined.

Numerical tables

Definition A numerical table for a limit at aa lists f(x)f(x) at inputs that approach aa from both sides, such as a±0.1a \pm 0.1, a±0.01a \pm 0.01 and a±0.001a \pm 0.001, and never at aa itself. Outputs that close in on one number from both sides suggest the limit. A calculator or computer rounds every number to a fixed count of digits, so at inputs extremely close to aa, subtracting two nearly equal numbers can wipe out the correct digits.

In plain words Come at aa from both sides in shrinking steps and watch the outputs. Stop before the inputs get so close that rounding takes over.

Example Here is a table for the function below near 00.

g(x)=x+9−3x.g(x) = \frac{\sqrt{x + 9} - 3}{x}.
xx−0.1-0.1−0.01-0.01−0.001-0.0010.0010.0010.010.010.10.1
g(x)g(x)0.1671320.1671320.1667130.1667130.1666710.1666710.1666620.1666620.1666200.1666200.1662060.166206

The outputs close in on about 0.166670.16667, which suggests 16\frac{1}{6}. Multiplying by the conjugate confirms it. For x≥−9x \ge -9 with x≠0x \ne 0,

g(x)=(x+9)−9x(x+9+3)=1x+9+3,\begin{aligned} g(x) &= \frac{(x + 9) - 9}{x\left(\sqrt{x + 9} + 3\right)} \\[0.8em] &= \frac{1}{\sqrt{x + 9} + 3}, \end{aligned}

which approaches 13+3=16\frac{1}{3 + 3} = \frac{1}{6}. Push the table too far, though, and a computer that keeps about 16 significant digits reports g(10−15)≈0.444g(10^{-15}) \approx 0.444 and g(10−16)=0g(10^{-16}) = 0. Those numbers come from rounding, not from the function.

Graph window

Definition A graph window, also called a viewing window, is the rectangle of the plane that a graphing tool shows, with xx from xmin⁡x_{\min} to xmax⁡x_{\max} and yy from ymin⁡y_{\min} to ymax⁡y_{\max}. The tool computes ff at finitely many sample inputs and joins the dots. So the picture can hide a hole or a single moved point, it can draw a false steep segment across a vertical asymptote, and it can turn fast oscillation into a pattern that is not there. A window centered at (a,L)(a, L) with half-width δ\delta and half-height ε\varepsilon pictures the limit condition. If every point of the graph with 0<∣x−a∣<δ0 < \lvert x - a \rvert < \delta lies strictly between the top and bottom edges, then those inputs give ∣f(x)−L∣<ε\lvert f(x) - L \rvert < \varepsilon.

In plain words Zoom in around the point where you expect the limit. If the curve crosses the window from its left edge to its right edge without escaping out the top or bottom, the outputs near aa stay near LL. Whatever the screen does not draw, it cannot show you.

Example In the window with 0≤x≤60 \le x \le 6 and 0≤y≤100 \le y \le 10 the graph of the opening function looks like the whole line y=x+3y = x + 3. The missing point (3,6)(3, 6) is a single point, far too small for any screen, and only the formula tells you it is missing. Now zoom to 2.99≤x≤3.012.99 \le x \le 3.01 and 5.99≤y≤6.015.99 \le y \le 6.01. The graph runs diagonally from the lower left corner to the upper right corner and never leaves through the top or bottom, which matches the limit 66.

Nonexistence

Definition The limit of ff at aa does not exist when no single real number LL meets the definition. There are three common ways this happens. In a jump, both one-sided limits exist but they differ. In unbounded behavior, the outputs on at least one side grow past every bound in size. In oscillation, there are two numbers p<qp < q such that every interval around aa, however small, contains inputs x≠ax \ne a with f(x)≤pf(x) \le p and inputs x≠ax \ne a with f(x)≥qf(x) \ge q. When the outputs grow without bound through positive values on both sides we write lim⁡x→af(x)=∞\lim_{x \to a} f(x) = \infty, but this only describes how the limit fails, since ∞\infty is not a real number. A two-sided limit also needs ff to be defined at nearby inputs on both sides, so at an endpoint of the domain only a one-sided limit can make sense.

In plain words A limit fails when the two sides disagree, when the outputs blow up, or when they keep swinging and never settle.

Example Three functions show the three failures at one point each.

  1. The function ∣x−1∣x−1\frac{\lvert x - 1 \rvert}{x - 1} equals −1-1 for every x<1x < 1 and 11 for every x>1x > 1. That is a jump, so there is no limit at 11.
  2. The function 1(x−1)2\frac{1}{(x - 1)^2} gives 1000010000 at both x=0.99x = 0.99 and x=1.01x = 1.01 and grows past every bound as xx approaches 11. Its graph has the vertical asymptote x=1x = 1, and there is no real limit.
  3. The function sin⁡(π/x)\sin(\pi / x) equals 11 at x=24k+1x = \frac{2}{4k + 1} and −1-1 at x=24k+3x = \frac{2}{4k + 3} for every positive whole number kk. Both kinds of input come as close to 00 as we like, so with p=−1p = -1 and q=1q = 1 the outputs oscillate and there is no limit at 00.

Evidence versus proof

Definition A table checks only finitely many inputs, and a graph shows only what its window and its sampling allow. Both give evidence and a good guess, never a proof. A proof is an argument that covers every input in some interval around aa with aa removed, usually algebra that rewrites f(x)f(x) for x≠ax \ne a in a form whose behavior near aa is clear, or an inequality that bounds ∣f(x)−L∣\lvert f(x) - L \rvert.

In plain words Use tables and graphs to guess. Then confirm the guess with an argument that works for all inputs near aa, not just the ones you tried.

Example Consider the function below near 00.

f(x)=x2x2+10−6.f(x) = \frac{x^2}{x^2 + 10^{-6}}.

It is even, so negative inputs give the same outputs as positive ones.

xx±0.5\pm 0.5±0.1\pm 0.1±0.05\pm 0.05±0.01\pm 0.01
f(x)f(x)0.9999960.9999960.99990.99990.99960.99960.9900990.990099

The table points to 11, but the limit is 00. For every x≠0x \ne 0 the denominator is larger than 10−610^{-6}, so

0≤f(x)≤x210−6=106x2,0 \le f(x) \le \frac{x^2}{10^{-6}} = 10^{6} x^2,

and the right side gets as small as we like as xx approaches 00. Inputs closer to 00 than the table went show the drop, with f(0.001)=0.5f(0.001) = 0.5 and f(0.0001)≈0.0099f(0.0001) \approx 0.0099.

Check yourself

  1. Make a table for the function below with x=1.9x = 1.9, 1.991.99, 2.012.01 and 2.12.1, guess its limit at 22, and then confirm the guess with algebra.
f(x)=x2+x−6x−2.f(x) = \frac{x^2 + x - 6}{x - 2}.
Show the answer

The table gives 4.94.9, 4.994.99, 5.015.01 and 5.15.1, which suggests 55. Since x2+x−6=(x−2)(x+3)x^2 + x - 6 = (x - 2)(x + 3), the function equals x+3x + 3 for every x≠2x \ne 2, and x+3x + 3 approaches 55. The limit is 55.

  1. Let f(x)=2x+1f(x) = 2x + 1 for x<3x < 3, f(x)=x2−2f(x) = x^2 - 2 for x>3x > 3 and f(3)=4f(3) = 4. Find both one-sided limits at 33, the two-sided limit and f(3)f(3).
Show the answer

From the left the outputs approach 2⋅3+1=72 \cdot 3 + 1 = 7, and from the right they approach 32−2=73^2 - 2 = 7. Both one-sided limits are 77, so the two-sided limit is 77. The value is f(3)=4f(3) = 4, which is different.

  1. For f(x)=1+sin⁡(π/x)f(x) = 1 + \sin(\pi / x) a table at x=0.1x = 0.1, 0.010.01 and 0.0010.001 gives exactly 11 every time. Does the limit at 00 exist?
Show the answer

No. Those inputs make π/x\pi / x a whole multiple of π\pi, where the sine is 00. At x=24k+1x = \frac{2}{4k + 1} the function equals 22, and at x=24k+3x = \frac{2}{4k + 3} it equals 00, for every positive whole number kk. These inputs come arbitrarily close to 00, so the outputs oscillate and there is no limit.

How this lesson connects

Where this leads in Calculus