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AP Calculus AB/BC Study Notes

1.2.2 Limit Notation

In calculus, the concept of limits helps us understand the behavior of functions as they approach a specific point. This approach is fundamental for dealing with functions at points where they may not be explicitly defined. The correct notation of limits is not just a formality but a crucial tool in calculus, offering a clear and concise way to express how functions behave near certain values.

Limit Notation

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Understanding Limit Notation

Limit notation is the language we use to describe the behavior of functions as variables approach specific values. It is a formal way of saying that as x gets closer and closer to some value cc, the function f(x)f(x) approaches some value R. The notation is:

limxcf(x)=R\lim_{x \to c} f(x) = R

This equation tells us that the limit of f(x)f(x), as xx approaches cc, is RR. This notation is pivotal in calculus for a few reasons:

  • Precision: It allows mathematicians to convey complex ideas precisely.
  • Clarity: It provides a clear way to indicate the point of approach and the expected value of the function at that point.
  • Versatility: It can describe limits from the left, right, or both, giving a complete picture of the function's behavior around c.

Worked Examples

Example 1: Linear Function Approach

Given f(x)=3x+2f(x) = 3x + 2, find the limit as xx approaches 4.

  • Expression:
limx4(3x+2)\lim_{x \to 4} (3x + 2)
  • Calculation:
=3(4)+2= 3(4) + 2
  • Result: 1414

The limit of f(x)f(x) as xx approaches 4 is 14.

Example 2: Handling Undefined Points

For f(x)=x21x1f(x) = \dfrac{x^2 - 1}{x - 1}, determine the limit as xx approaches 1.

  • Factor and Simplify:
f(x)=(x+1)(x1)x1f(x) = \dfrac{(x + 1)(x - 1)}{x - 1}
  • Cancel Common Terms:
f(x)=x+1f(x) = x + 1
  • Apply Limit:
limx1(x+1)=2\lim_{x \to 1} (x + 1) = 2

The limit is 2, demonstrating how limits navigate points where the function might not be directly defined.

Example 3: Infinite Limits

Evaluate limx21x2\lim_{x \to 2} \dfrac{1}{x - 2}.

  • Approach: As xx approaches 2, the denominator approaches 0, indicating the function approaches infinity.
  • Conclusion:
limx21x2=\lim_{x \to 2} \frac{1}{x - 2} = \infty

This shows how limit notation captures behavior leading to infinity.

Example 4: Trigonometric Limits

Find the limit of sin(x)\sin(x) as xx approaches π\pi.

  • Calculation:
limxπsin(x)=sin(π)\lim_{x \to \pi} \sin(x) = \sin(\pi)
  • Result: 00

Limits extend to trigonometric functions, showing how these functions behave near significant angles.

Practice Questions

To further understand the concept of limits and how to apply limit notation, work through the following questions. Attempt to solve them on your own before checking the solutions provided below.

Questions 1: Linear Function Limit

Evaluate limx3(5x7)\lim_{x \to 3} (5x - 7).

Questions 2: Quadratic Function Limit

Determine limx2(x24x+4)\lim_{x \to 2} (x^2 - 4x + 4).

Questions 3: Rational Function Limit

Calculate limx3x29x3\lim_{x \to 3} \frac{x^2 - 9}{x - 3}.

Solutions to Practice Questions

Solution to Question 1

For the limit limx3(5x7)\lim_{x \to 3} (5x - 7), follow these steps:

  • Step 1: Apply the limit directly by substituting x=3x = 3 into the equation.
5(3)75(3) - 7
  • Step 2: Simplify the result.
157=815 - 7 = 8

Therefore, the limit is 88, indicated by limx3(5x7)=8\lim_{x \to 3} (5x - 7) = 8.

Solution to Question 2

To find limx2(x24x+4)\lim_{x \to 2} (x^2 - 4x + 4):

  • Step 1: Substitute x=2x = 2 directly into the quadratic equation.
(2)24(2)+4(2)^2 - 4(2) + 4
  • Step 2: Carry out the arithmetic operations.
48+4=04 - 8 + 4 = 0

The limit of the quadratic function as xx approaches 22 is 00, shown by limx2(x24x+4)=0\lim_{x \to 2} (x^2 - 4x + 4) = 0.

Solution to Question 3

For the limit limx3x29x3\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}:

  • Step 1: Recognize that direct substitution would result in a (0/0) form. Factor the numerator to simplify.
(x+3)(x3)x3\frac{(x + 3)(x - 3)}{x - 3}
  • Step 2: Cancel the common term (x - 3).
x+3x + 3
  • Step 3: Now, apply the limit by substituting x=3x = 3 into the simplified function.
3+3=63 + 3 = 6

Thus, limx3x29x3=6\lim_{x \to 3} \frac{x^2 - 9}{x - 3} = 6, demonstrating the method to deal with expressions that simplify to a non-indeterminate form.

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