Education.com
Try
Brainzy
Try
Plus

Properties of Limits Help

By — McGraw-Hill Professional
Updated on Aug 31, 2011

Introduction to Properties of Limits

To increase our facility in manipulating limits, we have certain arithmetical and functional rules about limits. Any of these may be verified using the rigorous definition of limit that was provided at the beginning of the last section. We shall state the rules and get right to the examples.

If f and g are two functions, c is a real number, and lim xc f ( x ) and lim xc g ( x ) exist, then

Theorem 1

(a) lim xc ( f ± g )( x ) = lim xc f ( x ) ± lim xc g ( x );

(b) lim xc ( f · g ) ( x ) = (lim xc f ( x )) · (lim xc g ( x ));

(c) Foundations of Calculus 2.2 Properties of Limits

(d) lim xc ( α · f ( x )) = α · (lim xc f ( x )) for any constant α .

Some theoretical results, which will prove useful throughout our study of calculus, are these:

Theorem 2

Let a < c < b. A function f on the interval {x : a < x < b} cannot have two distinct limits at c .

Theorem 3

Foundations of Calculus 2.2 Properties of Limits

does not exist.

Theorem 4 (The Pinching Theorem)

Suppose that f, g, and h are functions whose domains each contain S = (a, c) ∪ (c, b). Assume further that

g ( x ) ≤ f ( x ) ≤ h ( x )

for all x Foundations of Calculus 2.2 Properties of Limits S. Refer to Fig. 2.4.

 

Foundations of Calculus 2.2 Properties of Limits

Fig. 2.4

If

Foundations of Calculus 2.2 Properties of Limits

 

and

Foundations of Calculus 2.2 Properties of Limits

then

Foundations of Calculus 2.2 Properties of Limits

Examples

Example 1

Calculate lim x → 3 4 x 3 − 7 x 2 + 5 x − 9.

Solution 1

We may apply Theorem 2.1(a) repeatedly to see that

Foundations of Calculus 2.2 Properties of Limits

We next observe that lim x → 3 x = 3. This assertion is self-evident, for when x is near to 3 then x is near to 3. Applying Theorem 2.1(d) and Theorem 2.1(b) repeatedly, we now see that

Foundations of Calculus 2.2 Properties of Limits

Of course lim x → 3 9 = 9.

Putting all this information into equation (*) gives

Foundations of Calculus 2.2 Properties of Limits

Example 2

Use the Pinching Theorem to analyze the limit

Foundations of Calculus 2.2 Properties of Limits

Solution 2

We observe that

−∣ x ∣ ≡ g ( x ) ≤ f ( x ) = x sin xh ( x ≡ ∣ x ∣.

Thus we may apply the Pinching Theorem. Obviously

Foundations of Calculus 2.2 Properties of Limits

We conclude that lim x → 0 f ( x ) = 0.

Example 3

Analyze the limit

Foundations of Calculus 2.2 Properties of Limits

Solution 3

The denominator tends to 0 while the numerator does not. According to Theorem 2.3, the limit cannot exist.

You Try It: Use the Pinching Theorem to calculate lim x → 0 x 2 sin x .

You Try It: What can you say about Foundations of Calculus 2.2 Properties of Limits

Find practice problems and solutions for these concepts at: Foundations of Calculus Practice Test.

Add your own comment

Ask a Question

Have questions about this article or topic? Ask
Ask
150 Characters allowed