Estimating Limits from Tables

Learn to estimate limit values by examining tables of function values

Using Tables to Find Limits

One of the most intuitive ways to understand limits is by creating a table of values and observing the pattern.

The Strategy

To estimate limā”xā†’af(x)\lim_{x \to a} f(x):

  1. Choose values approaching from the left (x < a)
  2. Choose values approaching from the right (x > a)
  3. Calculate f(x) for each value
  4. Look for a pattern - what value are the outputs approaching?

Example Setup

Let's find limā”xā†’2x2āˆ’4xāˆ’2\lim_{x \to 2} \frac{x^2 - 4}{x - 2} using a table.

Note: We can't just plug in x = 2 because we'd get 00\frac{0}{0} (undefined!)

From the left (x < 2):

| x | f(x) = x2āˆ’4xāˆ’2\frac{x^2 - 4}{x - 2} | |---|---| | 1.9 | (1.9)2āˆ’41.9āˆ’2=āˆ’0.39āˆ’0.1=3.9\frac{(1.9)^2 - 4}{1.9 - 2} = \frac{-0.39}{-0.1} = 3.9 | | 1.99 | (1.99)2āˆ’41.99āˆ’2=āˆ’0.0399āˆ’0.01=3.99\frac{(1.99)^2 - 4}{1.99 - 2} = \frac{-0.0399}{-0.01} = 3.99 | | 1.999 | (1.999)2āˆ’41.999āˆ’2ā‰ˆ3.999\frac{(1.999)^2 - 4}{1.999 - 2} \approx 3.999 |

From the right (x > 2):

| x | f(x) = x2āˆ’4xāˆ’2\frac{x^2 - 4}{x - 2} | |---|---| | 2.1 | (2.1)2āˆ’42.1āˆ’2=0.410.1=4.1\frac{(2.1)^2 - 4}{2.1 - 2} = \frac{0.41}{0.1} = 4.1 | | 2.01 | (2.01)2āˆ’42.01āˆ’2=0.04010.01=4.01\frac{(2.01)^2 - 4}{2.01 - 2} = \frac{0.0401}{0.01} = 4.01 | | 2.001 | (2.001)2āˆ’42.001āˆ’2ā‰ˆ4.001\frac{(2.001)^2 - 4}{2.001 - 2} \approx 4.001 |

The Pattern

Both sides are approaching 4!

Therefore: limā”xā†’2x2āˆ’4xāˆ’2=4\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4

Tips for Making Tables

  1. Start reasonably close - Use values like 0.1, 0.01 away from a
  2. Get progressively closer - Move to 0.01, 0.001, etc.
  3. Check both sides - Always examine left and right approaches
  4. Use a calculator - For complex functions, don't do it by hand
  5. Look for agreement - If both sides approach the same value, that's your limit!

When Tables Fail

Tables are estimates! They might be misleading if:

  • You don't get close enough to the point
  • The function oscillates wildly near the point
  • There's a very steep change right at the point

Tables are excellent for building intuition, but algebraic methods (coming in later lessons) are more precise.

šŸ“š Practice Problems

1Problem 1medium

ā“ Question:

Use a table to estimate limā”xā†’1xāˆ’1x2āˆ’1\lim_{x \to 1} \frac{x - 1}{x^2 - 1}. Check values from both sides.

šŸ’” Show Solution

Let's create a table approaching x = 1:

From the left (x < 1):

| x | xāˆ’1x2āˆ’1\frac{x - 1}{x^2 - 1} | Decimal | |---|---|---| | 0.9 | āˆ’0.1āˆ’0.19\frac{-0.1}{-0.19} | 0.526 | | 0.99 | āˆ’0.01āˆ’0.0199\frac{-0.01}{-0.0199} | 0.503 | | 0.999 | āˆ’0.001āˆ’0.001999\frac{-0.001}{-0.001999} | 0.500 |

From the right (x > 1):

| x | xāˆ’1x2āˆ’1\frac{x - 1}{x^2 - 1} | Decimal | |---|---|---| | 1.1 | 0.10.21\frac{0.1}{0.21} | 0.476 | | 1.01 | 0.010.0201\frac{0.01}{0.0201} | 0.498 | | 1.001 | 0.0010.002001\frac{0.001}{0.002001} | 0.500 |

Both sides approach 0.5, so:

limā”xā†’1xāˆ’1x2āˆ’1=0.5=12\lim_{x \to 1} \frac{x - 1}{x^2 - 1} = 0.5 = \frac{1}{2}

2Problem 2easy

ā“ Question:

Estimate lim(xā†’2) f(x) from the table: x: 1.9 1.99 1.999 2.001 2.01 2.1 f(x): 3.8 3.98 3.998 4.002 4.02 4.2

šŸ’” Show Solution

Step 1: Examine values as x approaches 2 from the left: x: 1.9 ā†’ 1.99 ā†’ 1.999 ā†’ 2 f(x): 3.8 ā†’ 3.98 ā†’ 3.998 ā†’ ? Values approach 4

Step 2: Examine values as x approaches 2 from the right: x: 2 ā† 2.001 ā† 2.01 ā† 2.1 f(x): ? ā† 4.002 ā† 4.02 ā† 4.2 Values approach 4

Step 3: Check agreement: Left-hand limit ā‰ˆ 4 Right-hand limit ā‰ˆ 4 Both agree!

Step 4: Conclusion: lim(xā†’2) f(x) = 4

Answer: 4

3Problem 3easy

ā“ Question:

Estimate lim(xā†’0) (sin(x)/x) from the table: x: 0.1 0.01 0.001 -0.001 -0.01 -0.1 f(x): 0.9983 0.99998 0.999999 0.999999 0.99998 0.9983

šŸ’” Show Solution

Step 1: Examine from the right (positive x): x: 0.1 ā†’ 0.01 ā†’ 0.001 ā†’ 0 f(x): 0.9983 ā†’ 0.99998 ā†’ 0.999999 ā†’ ? Approaching 1

Step 2: Examine from the left (negative x): x: -0.1 ā†’ -0.01 ā†’ -0.001 ā†’ 0 f(x): 0.9983 ā†’ 0.99998 ā†’ 0.999999 ā†’ ? Also approaching 1

Step 3: Verify symmetry: Function values are the same from both sides Both approach 1

Step 4: Estimate the limit: lim(xā†’0) (sin(x)/x) ā‰ˆ 1

Step 5: Note: This is a famous limit in calculus! Exact value is 1

Answer: 1

4Problem 4medium

ā“ Question:

Use a table to estimate lim(xā†’1) (xĀ³ - 1)/(x - 1)

šŸ’” Show Solution

Step 1: Create a table approaching from the left: x: 0.9 0.99 0.999 xĀ³-1: -0.271 -0.0299 -0.003 x-1: -0.1 -0.01 -0.001 f(x): 2.71 2.9701 2.997

Step 2: Create a table approaching from the right: x: 1.1 1.01 1.001 xĀ³-1: 0.331 0.0303 0.003 x-1: 0.1 0.01 0.001 f(x): 3.31 3.0301 3.003

Step 3: Analyze the pattern: From left: 2.71 ā†’ 2.9701 ā†’ 2.997 ā†’ ? From right: 3.31 ā†’ 3.0301 ā†’ 3.003 ā†’ ? Both approaching 3

Step 4: Estimate: lim(xā†’1) (xĀ³ - 1)/(x - 1) ā‰ˆ 3

Step 5: Verification (algebraic): xĀ³ - 1 = (x - 1)(xĀ² + x + 1) (xĀ³ - 1)/(x - 1) = xĀ² + x + 1 At x = 1: 1 + 1 + 1 = 3 āœ“

Answer: 3

5Problem 5hard

ā“ Question:

Estimate lim(xā†’0āŗ) (1/x) from the table. What happens? x: 0.1 0.01 0.001 0.0001 f(x): 10 100 1000 10000

šŸ’” Show Solution

Step 1: Examine the pattern: As x gets smaller (closer to 0): x: 0.1 ā†’ 0.01 ā†’ 0.001 ā†’ 0.0001 ā†’ 0āŗ f(x): 10 ā†’ 100 ā†’ 1000 ā†’ 10000 ā†’ ?

Step 2: Observe the behavior: The function values are growing without bound They're increasing toward infinity

Step 3: Conclusion: The limit does not exist as a finite number We say: lim(xā†’0āŗ) (1/x) = āˆž

Step 4: Important note: This is an "infinite limit" āˆž is not a number, it describes unbounded growth

Step 5: Graphical interpretation: The graph has a vertical asymptote at x = 0

Answer: The limit is āˆž (infinite limit)

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