Calculus 2.3 Calculating Limits Using the Limit Laws

Introduction to Limit Laws

• Moving from informal methods of finding limits, such as tables and graphs, to formal methods using limit laws.

Limit Laws Overview

• Law 1: Sum Law

  • If ( c ) is a constant, then:[ \lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) ]

• Law 2: Difference Law

  • Similar to the sum law:[ \lim_{x \to a} (f(x) - g(x)) = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) ]

• Law 3: Constant Factor Law

  • A constant can be factored out:[ \lim_{x \to a} (c f(x)) = c \lim_{x \to a} f(x) ]

• Law 4: Product Law

  • The limit of a product can be calculated by multiplying the limits:[ \lim_{x \to a} (f(x) g(x)) = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) ]

• Law 5: Quotient Law

  • The limit of a quotient can be computed by dividing the limits, provided the limit of the denominator is not zero:[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \quad (g(a)
    eq 0) ]

Evaluating Limits with Limit Laws

Example 1: Limit of f(x) + 5g(x) as x approaches -2

• Split the limit:[ \lim_{x \to -2} (f(x) + 5g(x)) = \lim_{x \to -2} f(x) + 5 \lim_{x \to -2} g(x) ]

• Using a graph:

  • ( \lim_{x \to -2} f(x) = 1 )

  • ( \lim_{x \to -2} g(x) = -1 )

• Thus:[ 1 + 5(-1) = 1 - 5 = -4 ]

Example 2: Limit of f(x)g(x) as x approaches 1

• Check limits:

  • ( g(x) ) does not exist as left and right limits are not equal.

• Left Handed Limit:

  • ( \lim_{x \to 1^-} f(x) = 2, \lim_{x \to 1^-} g(x) = -2 )

  • Product: ( 2 imes -2 = -4 )

• Right Handed Limit:

  • ( \lim_{x \to 1^+} g(x) = -1 )

  • Product: ( 2 imes -1 = -2 )

• Since limits do not match, overall limit does not exist.

Example 3: Limit of f(x)/g(x) as x approaches 2

• Find limits:

  • ( \lim_{x \to 2} f(x) = 1.4 )

  • ( \lim_{x \to 2} g(x) = 0 )

• Conclusion: The limit does not exist due to nonzero numerator and zero denominator.

Power and Root Laws

• Power Law: [ \lim_{x \to a} (f(x)^n) = (\lim_{x \to a} f(x))^n ]

• Roots Law:

  • For nth roots:[ \lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)} ]

• Justify limits using these laws through direct substitution.

Direct Substitution Property

• For polynomials or rational functions, ( \lim_{x \to A} f(x) = f(A) ) if A is in the domain.

Example 4: Limit as x approaches 1 for ( \frac{x^2 - 1}{x - 1} )

• Factor the expression:[ \lim_{x \to 1} \frac{(x - 1)(x + 1)}{x - 1} \rightarrow \lim_{x \to 1} (x + 1) \rightarrow 2 ]

Piecewise Functions and Limits

• If ( f = g ) when ( x
  eq A ), then ( \lim_{x \to A} f(x) = \lim_{x \to A} g(x) ) when limits exist.

Example 5: Limit of g(x) at x = 1

• ( g(x) = x + 1 ) for ( x
  eq 1 );

• Thus:[ \lim_{x \to 1} g(x) = 2 ]

Greatest Integer Function

• Defined by the greatest integer less than or equal to ( x ) (floor function).

• Analyzing limits using left and right-sided limits.

Example 6: ( \lim_{x \to 3} \text{greatest integer function} )

• Left:

  • Approaching values like 2.9 gives limit 2.

• Right:

  • Approaching values like 3.1 gives limit 3.

• Conclusion: Limit does not exist as left and right limits do not match.

Squeeze Theorem

• If ( f(x) \leq g(x) \leq h(x) ) and ( \lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L ), then ( \lim_{x \to a} g(x) = L ).

Example 7: Showing ( \lim_{x \to 0} (x^2 \sin(1/x)) = 0 )

• Squeeze:

  • |sin(1/x)| is always between -1 and 1.

• Hence:[ -x^2 \leq x^2 \sin(1/x) \leq x^2 ]

  • As x approaches 0, both bounds approach 0, thus limit is 0.