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Estimating Limits Numerically and Graphically

MATH161
Reference: Stewart 2.2  •  Chapter: 1  •  Section: 2

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Estimating Limits Numerically and Graphically

Before We Can Calculate, We Must Observe

Before you learn algebraic techniques for computing limits exactly, you need to develop intuition for what limits "look like." This skill teaches you to estimate limits by:

  1. Building a table of values that approach the target from both sides
  2. Reading limit behavior from graphs by tracing along curves

These skills remain valuable even after you learn algebraic methods—they serve as sanity checks and help you understand functions that don't have closed-form limits.

Warning: Numerical and graphical methods can be deceived by tricky functions. This skill also teaches you what can go wrong.

Prerequisite Map

This skill
Estimating Limits
Unlocks
Limit LawsAlgebraic Evaluation

Quick Reference

Property Value
Concept Limits
Course MATH161
Section Stewart 1.5
Difficulty Beginner
Time ~15 minutes

Key Concepts

The Numerical Approach: Building a Table

To estimate $\lim_{x \to a} f(x)$ numerically:

Step 1: Choose values of $x$ approaching $a$ from both sides:

Step 2: Compute $f(x)$ for each value.

Step 3: Look for a pattern—are the values converging to a single number?

Example: Estimating $\lim_{x \to 1} \frac{x-1}{x^2-1}$

$x$ $\frac{x-1}{x^2-1}$
0.9 0.5263...
0.99 0.5025...
0.999 0.5003...
1.001 0.4998...
1.01 0.4975...
1.1 0.4762...

The values are clearly converging to $0.5$ from both sides. We estimate: $$\lim_{x \to 1} \frac{x-1}{x^2-1} \approx 0.5$$

(The exact value is $\frac{1}{2}$, which we can verify by factoring: $\frac{x-1}{(x-1)(x+1)} = \frac{1}{x+1} \to \frac{1}{2}$.)

The Graphical Approach: Reading from Curves

To estimate $\lim_{x \to a} f(x)$ from a graph:

Step 1: Locate $x = a$ on the horizontal axis.

Step 2: Trace along the curve from both sides toward $x = a$.

Step 3: Observe what $y$-value the curve approaches.

Key points:

Graphical Estimation Diagram

    y
    |
  2 +   ┌─○        ← Open circle: curve approaches y=2
    |   │
  1 +───┘ ●        ← Filled dot: f(a) = 1, but that's not the limit!
    |
  0 +────+────→ x
         a

Limit = 2 (where the curve is heading)
f(a) = 1 (where the point actually is)

Common Pitfalls in Estimation

Pitfall Example How to Avoid
Calculator rounding $\frac{\sin x}{x}$ near $x = 10^{-15}$ may show 0 Don't go too close; values like $0.001$ are fine
Oscillating functions $\sin(\frac{1}{x})$ near $x = 0$ Make a table; oscillation will be visible
Confusing $f(a)$ with limit Piecewise functions Check the curve trend, not the isolated point
One-sided difference $\frac{\|x\|}{x}$ near $x = 0$ Always check BOTH sides

The Oscillation Problem

Some functions oscillate infinitely near a point. For example:

$$f(x) = \sin\left(\frac{1}{x}\right)$$

As $x \to 0$, the argument $\frac{1}{x} \to \pm\infty$, causing sine to oscillate between $-1$ and $1$ infinitely often.

$x$ $\sin(1/x)$
0.1 $-0.544$
0.05 $0.912$
0.02 $0.455$
0.01 $-0.506$
0.005 $-0.959$
0.002 $-0.879$

The values don't settle down! This indicates $\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$ does not exist.

When to Trust Your Estimate

Your numerical estimate is reliable when:

Your estimate may be unreliable when:

Practice Problems

Level 1 Reading from a Table

Based on the following table, estimate $\lim_{x \to 0} f(x)$:

$x$ $-0.1$ $-0.01$ $-0.001$ $0.001$ $0.01$ $0.1$
$f(x)$ $2.87$ $2.9987$ $2.999987$ $3.000013$ $3.0013$ $3.13$
Thought Process

Look at the pattern in the $f(x)$ row. As $x$ gets closer to $0$ (from either side), where are the values heading?

Show Answer

From the left: $2.87 \to 2.9987 \to 2.999987 \to \ldots$ approaching $3$

From the right: $3.13 \to 3.0013 \to 3.000013 \to \ldots$ approaching $3$

Both sides converge to the same value, so:

$$\lim_{x \to 0} f(x) = 3$$

Level 2 Creating Your Own Table

Use a table of values to estimate $\lim_{x \to 0} \frac{e^x - 1}{x}$.

(Use at least 3 values from each side.)

Thought Process

Direct substitution gives $\frac{0}{0}$ (indeterminate). So we need to build a table with values like $x = \pm 0.1, \pm 0.01, \pm 0.001$ and compute $\frac{e^x - 1}{x}$ for each.

Show Answer
$x$ $e^x - 1$ $\frac{e^x - 1}{x}$
$-0.1$ $-0.0952$ $0.9516$
$-0.01$ $-0.00995$ $0.9950$
$-0.001$ $-0.000999$ $0.9995$
$0.001$ $0.001001$ $1.0005$
$0.01$ $0.01005$ $1.0050$
$0.1$ $0.1052$ $1.0517$

From left: $0.9516 \to 0.9950 \to 0.9995 \to \ldots$ approaching $1$

From right: $1.0517 \to 1.0050 \to 1.0005 \to \ldots$ approaching $1$

$$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$

Note: This is an important limit that appears in the definition of the derivative of $e^x$.

Level 3 Reading Limits from a Graph

The graph of $g(x)$ is shown below. Find all requested limits.

    y
    |
  4 +            /
    |           /
  3 +   ●──────○
    |  /
  2 + /
    |/
  1 +
    |    ●
  0 +----+----+----+----→ x
         1    2    3    4

Key features:

  • At $x = 1$: filled point at $y = 3$
  • At $x = 2$: open circle at $y = 3$, separate filled point at $y = 0$
  • At $x = 3$: curve continues upward

Find:

  1. $\lim_{x \to 1} g(x)$
  2. $\lim_{x \to 2^-} g(x)$
  3. $\lim_{x \to 2^+} g(x)$
  4. $\lim_{x \to 2} g(x)$
  5. $g(2)$
Thought Process

Trace along the curve:

  • Near $x = 1$: the curve and the point agree
  • Near $x = 2$: trace from left (along the horizontal line at $y = 3$), trace from right (along the rising line), and note where the filled point is
Show Answer

(a) $\lim_{x \to 1} g(x) = 3$ The curve smoothly passes through the point $(1, 3)$.

(b) $\lim_{x \to 2^-} g(x) = 3$ Approaching from the left, we follow the horizontal line at $y = 3$.

(c) $\lim_{x \to 2^+} g(x) = 3$ Approaching from the right, we follow the rising curve down toward the open circle at $y = 3$.

(d) $\lim_{x \to 2} g(x) = 3$ Both one-sided limits equal $3$, so the two-sided limit exists and equals $3$.

(e) $g(2) = 0$ The filled point (actual function value) is at $y = 0$.

Note: The limit exists and equals $3$, but the function value is $0$. This is a removable discontinuity.

Level 4 Detecting a Non-Existent Limit

Use a table of values to investigate $\lim_{x \to 0} \sin\left(\frac{\pi}{x}\right)$.

Does the limit exist? Explain your reasoning based on the numerical evidence.

Thought Process

As $x \to 0$, the argument $\frac{\pi}{x}$ becomes very large (positive or negative). The sine function oscillates between $-1$ and $1$. Build a table with values like $x = 0.5, 0.25, 0.1, 0.05, \ldots$ and see if there's a pattern.

Show Answer
$x$ $\frac{\pi}{x}$ $\sin\left(\frac{\pi}{x}\right)$
$1$ $\pi$ $0$
$0.5$ $2\pi$ $0$
$0.4$ $2.5\pi$ $-1$
$0.333...$ $3\pi$ $0$
$0.25$ $4\pi$ $0$
$0.2$ $5\pi$ $0$
$0.1$ $10\pi$ $0$
$0.09$ $\approx 34.9$ $\approx -0.96$
$0.08$ $\approx 39.3$ $\approx 0.91$

Observation: The values don't settle to any single number. We get $0$'s at certain points (when $\frac{\pi}{x}$ is a multiple of $\pi$), but between those, the function takes values across $[-1, 1]$.

Conclusion: The limit does not exist.

As $x \to 0$, the function oscillates infinitely many times between $-1$ and $1$, never settling to a single value. This is different from a jump discontinuity—here the function oscillates infinitely often near the point.

Level 5 The Calculator Deception

Consider $f(x) = x \cdot \sin\left(\frac{1}{x}\right)$ for $x \neq 0$.

  1. Use a table of values to estimate $\lim_{x \to 0} f(x)$.
  2. Compare with $g(x) = \sin\left(\frac{1}{x}\right)$. Why does $f(x)$ have a limit while $g(x)$ doesn't?
  3. What theorem from later sections will confirm your numerical estimate?
Thought Process

For part (a), even though $\sin\left(\frac{1}{x}\right)$ oscillates, the factor of $x$ in front becomes very small. What happens when you multiply a bounded oscillation by something approaching zero?

For part (b), think about what the extra factor of $x$ does to the oscillating part.

For part (c), there's a theorem about limits that involves bounding a function between two others...

Show Answer

(a) Numerical estimation:

$x$ $\sin(1/x)$ $x \cdot \sin(1/x)$
$0.1$ $-0.544$ $-0.0544$
$0.05$ $0.912$ $0.0456$
$0.01$ $-0.506$ $-0.00506$
$0.005$ $-0.959$ $-0.00479$
$0.001$ $0.827$ $0.000827$

Despite the oscillating sine values, the product $x \cdot \sin(1/x)$ is getting closer to $0$!

$$\lim_{x \to 0} x \cdot \sin\left(\frac{1}{x}\right) = 0$$

(b) Why $f$ has a limit but $g$ doesn't:

For $g(x) = \sin\left(\frac{1}{x}\right)$: The oscillation amplitude stays constant at $1$. No matter how close $x$ gets to $0$, the function swings between $-1$ and $1$.

For $f(x) = x \cdot \sin\left(\frac{1}{x}\right)$: The oscillation still happens, BUT it's being multiplied by $x$, which approaches $0$. So the "height" of the oscillation shrinks to $0$ as $x \to 0$.

Think of it as a sine wave whose amplitude is $\vert x\vert $—and that amplitude goes to $0$.

(c) The Squeeze Theorem (coming in Section 1.6):

Since $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$, we have:

$$-\vert x\vert \leq x \cdot \sin\left(\frac{1}{x}\right) \leq \vert x\vert $$

Since $\lim_{x \to 0} (-\vert x\vert ) = 0$ and $\lim_{x \to 0} \vert x\vert = 0$, the Squeeze Theorem guarantees:

$$\lim_{x \to 0} x \cdot \sin\left(\frac{1}{x}\right) = 0$$

Mastery Checklist

Mental Model

The Detective Approach: Estimating limits is like detective work. You're gathering evidence (table values, graph traces) to determine where the function is "trying to go." But remember: evidence can be misleading. Oscillating functions can fool you, and calculator precision has limits. Always check both sides, and use numerical estimates as a guide to be confirmed by algebraic methods later.


Connections

Looking back:

Looking ahead:

Real-world connections:


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Last updated: 2026-01-22