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AP Calculus AB/BC Study Notes

1.2.1 Understanding Limits

Limits form the cornerstone of calculus, embodying the value a function approaches as its input nears a specific point. This concept is pivotal in defining function behavior at particular points, especially where the function might not be explicitly defined.

The Concept of a Limit

  • Defines the behavior of f(x)f(x) as xx approaches a point cc, aiming for a real number RR.
  • Crucial for analyzing functions at undefined points or exhibiting unusual behavior.
Graph of Limits

Image courtesy of BYJUS

Worked Examples

Example 1: Direct Substitution

Find limx4(3x+2)\lim_{x \to 4} (3x + 2)

Substitute xx with 4:

limx4(3x+2)=3(4)+2=12+2=14\lim_{x \to 4} (3x + 2) = 3(4) + 2 = 12 + 2 = 14

Example 2: Factoring and Simplifying

Find limx1x21x1\lim_{x \to 1} \dfrac{x^2 - 1}{x - 1}

1. Factor the numerator:

x21x1=(x+1)(x1)x1\dfrac{x^2 - 1}{x - 1} = \dfrac{(x + 1)(x - 1)}{x - 1}

2. Cancel common terms:

(x+1)(x1)x1=x+1\dfrac{(x + 1)(x - 1)}{x - 1} = x + 1

3. Substitute xx with 1:

limx1(x+1)=1+1=2\lim_{x \to 1} (x + 1) = 1 + 1 = 2

Example 3: Graphical Interpretation

Estimate limx0+1x\lim_{x \to 0^+} \dfrac{1}{x}.

Graph of f(x) = 1/x

1. As xx approaches 0 from the right, 1x\frac{1}{x} grows infinitely large.

2. The function does not approach a finite number but instead goes towards infinity:

limx0+1x=\lim_{x \to 0^+} \dfrac{1}{x} = \infty

Importance of Limits

  • Foundation for Calculus: Enables understanding of derivatives, integrals.
  • Defines Continuity: A function is continuous at a point if the function's limit at that point equals its value.
  • Analyzes Asymptotic Behavior: Crucial for examining function behavior near undefined points or as functions approach infinity.

Practice Questions

Question 1

Find limx3(2x25x+1)\lim_{x \to 3} (2x^2 - 5x + 1).

Question 2

Calculate limx2x2+3x+2x+2\lim_{x \to -2} \dfrac{x^2 + 3x + 2}{x + 2}.

Question 3

Determine limx0sin(x)x\lim_{x \to 0} \dfrac{\sin(x)}{x}.

Solutions to Practice Questions

Solution to Question 1

Find limx3(2x25x+1)\lim_{x \to 3} (2x^2 - 5x + 1).

  • Step 1: Substitute x=3x = 3 into the function.
limx3(2x25x+1)=2(3)25(3)+1\lim_{x \to 3} (2x^2 - 5x + 1) = 2(3)^2 - 5(3) + 1
  • Step 2: Perform the calculation.
=1815+1=4= 18 - 15 + 1 = 4
  • Conclusion:
limx3(2x25x+1)=4\lim_{x \to 3} (2x^2 - 5x + 1) = 4

Solution to Question 2

Calculate limx2x2+3x+2x+2\lim_{x \to -2} \dfrac{x^2 + 3x + 2}{x + 2}.

  • Step 1: Factor the numerator and simplify the expression.
x2+3x+2x+2=(x+1)(x+2)x+2\dfrac{x^2 + 3x + 2}{x + 2} = \dfrac{(x + 1)(x + 2)}{x + 2}
  • Step 2: Cancel the common factor (x+2)(x + 2).
=x+1= x + 1
  • Step 3: Substitute x=2x = -2 into the simplified function.
limx2(x+1)=2+1=1\lim_{x \to -2} (x + 1) = -2 + 1 = -1
  • Conclusion:
limx2x2+3x+2x+2=1\lim_{x \to -2} \dfrac{x^2 + 3x + 2}{x + 2} = -1

Solution to Question 3

Determine limx0sin(x)x\lim_{x \to 0} \dfrac{\sin(x)}{x}.

  • Recognize this as a fundamental limit in calculus.
limx0sin(x)x=1\lim_{x \to 0} \dfrac{\sin(x)}{x} = 1
  • Conclusion:
limx0sin(x)x=1\lim_{x \to 0} \dfrac{\sin(x)}{x} = 1

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