Lecture 2.3
Limits and Derivatives
2.3 Calculating Limits Using the Limit Laws
Properties of Limits
4 Properties of Limits (1 of 5)
In this section, we utilize several properties of limits, referred to as the Limit Laws, to compute limits.
Limit Laws
Suppose that c is a constant and the limits ( \lim{x \to a} f(x) ) and ( \lim{x \to a} g(x) ) exist. Then:
( \lim{x \to a} (f(x) + g(x)) = \lim{x \to a} f(x) + \lim_{x \to a} g(x) ) (Sum Law)
( \lim{x \to a} (f(x) - g(x)) = \lim{x \to a} f(x) - \lim_{x \to a} g(x) ) (Difference Law)
( \lim{x \to a} (c f(x)) = c \lim{x \to a} f(x) ) (Constant Multiple Law)
( \lim{x \to a} (f(x) g(x)) = \lim{x \to a} f(x) \cdot \lim_{x \to a} g(x) ) (Product Law)
( \lim{x \to a} \left( \frac{f(x)}{g(x)} \right) = \frac{\lim{x \to a} f(x)}{\lim{x \to a} g(x)} ), provided ( \lim{x \to a} g(x) \neq 0 ) (Quotient Law)
5 Properties of Limits (2 of 5)
The five laws can be articulated verbally as follows:
Sum Law: The limit of a sum is the sum of the limits.
Difference Law: The limit of a difference is the difference of the limits.
Constant Multiple Law: The limit of a constant times a function is the constant times the limit of the function.
6 Properties of Limits (3 of 5)
Further definitions:
Product Law: The limit of a product is the product of the limits.
Quotient Law: The limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero.
Example: If ( f(x) ) approaches ( L ) and ( g(x) ) approaches ( M ), it follows that ( f(x) + g(x) ) approaches ( L + M ).
Example 1: Evaluating Limits Using Limit Laws
(a) Compute ( \lim_{x \to -2} (f(x) + g(x)) )
From the graphs of ( f ) and ( g ):
( \lim_{x \to -2} f(x) = -2 )
( \lim_{x \to -2} g(x) = -1 )
By applying Limit Laws:
( \lim{x \to -2} (f(x) + g(x)) = \lim{x \to -2} f(x) + \lim_{x \to -2} g(x) = -2 + (-1) = -3 )
(b) Compute ( \lim_{x \to 1} (f(x) g(x)) )
Observations:
( \lim_{x \to 1} f(x) = 1 )
( \lim_{x \to 1} g(x) ) does not exist (left and right limits differ).
Cannot use Law 4 because ( g(x) ) does not satisfy the limit existence requirements.
Conclusion: ( \lim_{x \to 1} (f(x) g(x)) ) does not exist.
(c) Compute ( \lim_{x \to 2} \frac{f(x)}{g(x)} )
From the graphs:
( \lim_{x \to 2} f(x) = 1.4 )
( \lim_{x \to 2} g(x) = 0 )
The limit does not exist since the denominator approaches zero while the numerator approaches a finite number.
7 Properties of Limits (4 of 5)
Using the Product Law repeatedly with ( g(x) = f(x) ) yields:
Power Law: ( \lim{x \to a} (f(x))^n = (\lim{x \to a} f(x))^n ) where ( n ) is a positive integer.
Root Law: ( \lim{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim{x \to a} f(x)} ) where ( n ) is a positive integer and ( f(x) ) is assumed to be non-negative.
Evaluating Limits by Direct Substitution
14 Evaluating Limits by Direct Substitution (1 of 2)
Direct Substitution Property: If ( f ) is a polynomial or a rational function and ( a ) is in the domain of ( f ), then ( \lim_{x \to a} f(x) = f(a) ).
Functions exhibiting the Direct Substitution Property are identified as continuous at ( a ).
15 Example 3: Evaluating a Limit by Direct Substitution
Problem: Find ( \lim_{x \to 1} \frac{x^2 - 1}{x - 1} ).
Notice:
Direct substitution ( x = 1 ) leads to an undefined expression (0/0).
Alternative: Factor the numerator as a difference of squares: ( x^2 - 1 = (x - 1)(x + 1) ).
16 Example 3 – Solution
Cancel common factor ( (x - 1) ) and compute limit as follows:
( \lim_{x \to 1} (x + 1) = 1 + 1 = 2 ).
17 Evaluating Limits by Direct Substitution (2 of 2)
General principle:
If ( f(x) = g(x) ) for ( x \neq a ), then ( \lim{x \to a} f(x) = \lim{x \to a} g(x) ), provided the limits exist.
Using One-Sided Limits
19 Using One-Sided Limits (1 of 1)
Some limits are most accurately found by calculating left-hand and right-hand limits first.
Theorem: A two-sided limit exists if and only if both one-sided limits exist and are equal:
( \lim{x \to a} f(x) = L ) if and only if ( \lim{x \to a^-} f(x) = L ) and ( \lim_{x \to a^+} f(x) = L ).
Example 7: Evaluating One-Sided Limits
Compute ( \lim_{x \to 0} x ).
For ( x > 0 ), ( f(x) = x ) approaches 0.
For ( x < 0 ), ( f(x) = -x ) approaches 0.
By defining limits:
( \lim_{x \to 0^+} f(x) = 0 )
( \lim_{x \to 0^-} f(x) = 0 )
Conclude: ( \lim_{x \to 0} x = 0 ).
The Squeeze Theorem
23 The Squeeze Theorem (1 of 2)
Theorems concerning limits of functions with respective inequalities:
Theorem: If ( f(x) \leq g(x) ) near ( a ) (except possibly at ( a )), and limits of both ( f ) and ( g ) exist, then ( \lim{x \to a} f(x) \leq \lim{x \to a} g(x) ).
Squeeze Theorem: If ( f(x) \leq g(x) \leq h(x) ) near ( a ) (except possibly at ( a )), and ( \lim{x \to a} f(x) = \lim{x \to a} h(x) = L ), then ( \lim_{x \to a} g(x) = L ).
24 The Squeeze Theorem (2 of 2)
The Squeeze Theorem (also known as the Sandwich Theorem or Pinching Theorem) assures that if ( g(x) ) is bounded by both ( f(x) ) and ( h(x) ) close to ( a ), and if both ( f ) and ( h ) converge to the same limit ( L ) at ( a ), then ( g ) must also converge to ( L ) as ( x \to a ).
25 Example 11: Using the Squeeze Theorem
Show that ( \lim_{x \,to \; 0} \frac{ \sin x}{x} = 1 ).
Note: Cannot rewrite limit as product of individual limits since ( \lim_{x \to 0} \sin(1/x) ) does not exist.
We will utilize the Squeeze Theorem instead.
26 Example 11 – Solution (1 of 3)
To apply the Squeeze Theorem, find functions ( f ) and ( h ) such that:
( f(x) = -x^2 ) and ( h(x) = x^2 ) satisfy ( -x^2 \, \leq \, x^2 \sin(1/x) \, \leq \, x^2 ) as both approach ( 0 ) as ( x \to 0 ).
27 Example 11 – Solution (2 of 3)
Multiply inequalities by the positive function ( 2x ):
Results: ( -2x^3 \leq x^2 \sin(1/x) \, \leq 2x^3 ).
As ( x \to 0 ), both bounds approach 0.
28 Example 11 – Solution (3 of 3)
Hence, applying the Squeeze Theorem yields:
( \lim_{x \,to \; 0} \frac{
\sin x}{x} = 1 ).