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IB DP Maths AA HL Study Notes

5.7.1 Evaluating Limits

Evaluating limits is a cornerstone of calculus, providing insight into the behaviour of functions as they approach specific points. In this section, we delve into three primary techniques for evaluating limits: direct substitution, factorisation, and rationalisation. Understanding these techniques requires a solid grasp of basic mathematical concepts, such as the properties of logarithms, which can influence how we approach complex limits.

Direct Substitution

Direct substitution is the most straightforward method for evaluating limits. It involves plugging the value directly into the function.

Process:

1. Substitute the value of the variable for which the limit is being taken directly into the function.

2. If the result is a defined value, that is the limit.

3. If the result is an indeterminate form (like 0/0), then other methods like factorisation or rationalisation might be needed. In cases involving rational functions, a review of the properties of rational functions might provide additional insights for handling such indeterminate forms.

Example 1:

Evaluate the limit of (x2 + 3x - 4) as x approaches 2.

Solution:

Using direct substitution, we substitute x = 2 into the function:

(22 + 3(2) - 4) = 6

Thus, the limit of (x2 + 3x - 4) as x approaches 2 is 6.

Example 2:

Evaluate the limit of (2x + 7) as x approaches 5.

Solution:

Using direct substitution:

2(5) + 7 = 17

Thus, the limit of (2x + 7) as x approaches 5 is 17.

Factorisation

Factorisation is the process of expressing the function as a product of its factors. This method becomes particularly useful when direct substitution results in an indeterminate form.

Process:

1. Factorise the function.

2. Cancel out common terms if possible.

3. Use direct substitution on the simplified function.

For more complex functions, understanding basic differentiation rules can aid in the factorisation process by identifying potential simplifications.

Example 1:

Evaluate the limit of (x2 - 9)/(x - 3) as x approaches 3.

Solution:

Factorising the numerator, we get:

(x + 3)(x - 3)/(x - 3)

Cancelling out the common term (x - 3):

x + 3

Now, using direct substitution:

The limit of (x + 3) as x approaches 3 is 6.

Thus, the limit of (x2 - 9)/(x - 3) as x approaches 3 is 6.

Example 2:

Evaluate the limit of (x2 - 16)/(x - 4) as x approaches 4.

Solution:

Factorising the numerator:

(x + 4)(x - 4)/(x - 4)

Cancelling out the common term (x - 4):

x + 4

Using direct substitution:

The limit of (x + 4) as x approaches 4 is 8.

Thus, the limit of (x2 - 16)/(x - 4) as x approaches 4 is 8.

Rationalisation

Rationalisation is a technique used to evaluate limits involving roots. It involves multiplying the function by a conjugate to eliminate the root from the denominator.

Process:

1. Identify the conjugate of the function.

2. Multiply the function by its conjugate.

3. Simplify the function.

4. Use direct substitution.

Applying basic integration techniques alongside rationalisation can be particularly effective for limits involving complex root structures.

Example 1:

Evaluate the limit of (sqrt(x) - 1)/(x - 1) as x approaches 1.

Solution:

Multiplying by the conjugate:

(sqrt(x) - 1)(sqrt(x) + 1)/(x - 1)(sqrt(x) + 1)

This results in:

x - 1

Now, using direct substitution:

The limit of (sqrt(x) + 1) as x approaches 1 is 1/2.

Thus, the limit of (sqrt(x) - 1)/(x - 1) as x approaches 1 is 1/2.

Example 2:

Evaluate the limit of (sqrt(x) - 2)/(x - 4) as x approaches 4.

Solution:

Multiplying by the conjugate:

(sqrt(x) - 2)(sqrt(x) + 2)/(x - 4)(sqrt(x) + 2)

This results in:

x - 4

Using direct substitution:

The limit of (sqrt(x) + 2) as x approaches 4 is 1/4.

Thus, the limit of (sqrt(x) - 2)/(x - 4) as x approaches 4 is 1/4.

In instances where direct substitution after rationalisation leads to complex limits, revisiting concepts such as L'Hôpital's Rule can offer a path forward by applying differentiation to both the numerator and the denominator, thereby simplifying the limit evaluation process.

Understanding the principles behind evaluating limits is fundamental in calculus and extends into various applications across mathematics. As students progress, exploring additional resources and topics within IB Maths, such as those mentioned, will enrich their learning experience and provide a more comprehensive understanding of mathematical concepts.

FAQ

Factorisation involves expressing a function as a product of its factors. When evaluating limits, factorisation can help simplify the function, especially when direct substitution results in an indeterminate form. By factorising, we can often cancel out terms in the numerator and denominator that lead to the indeterminate form. Once the function is simplified, direct substitution can then be applied to evaluate the limit. Factorisation is a versatile tool that can be used in conjunction with other methods, like rationalisation, to tackle more complex limit problems.

Rationalisation is a technique used to eliminate square roots from the denominator of a fraction. It involves multiplying the numerator and denominator by the conjugate of the denominator. The conjugate of a binomial expression a + sqrt(b) is a - sqrt(b), and vice versa. When you multiply these conjugates, the square root terms cancel out, leaving a rational expression. This new expression often allows for easier evaluation of the limit, especially when direct substitution initially results in an indeterminate form. Rationalisation is particularly useful for functions involving the difference of square roots.

While direct substitution, factorisation, and rationalisation are powerful techniques, they are not exhaustive. There are limits that cannot be evaluated using just these methods. Some functions may require more advanced techniques, such as L'Hopital's Rule, or a combination of multiple methods. Additionally, there are limits that do not exist, meaning the function doesn't approach a specific value as the variable approaches a particular point. It's essential to have a comprehensive toolkit of methods and to recognise when a particular technique is not suitable for evaluating a given limit.

Indeterminate forms are expressions whose values are not immediately apparent or well-defined. Common examples include 0/0, infinity/infinity, 0 * infinity, infinity - infinity, 00, infinity0, and 1infinity. These forms are problematic because they don't give a clear indication of the function's behaviour as it approaches the limit. For instance, 0/0 could represent any real number, depending on the function. When faced with an indeterminate form, it's a sign that further manipulation or alternative methods are needed to evaluate the limit accurately.

Direct substitution is the simplest method to evaluate limits, but it's not always applicable. When you substitute the value into the function and get an indeterminate form, like 0/0 or infinity/infinity, it means the function is not defined at that point. In such cases, direct substitution doesn't provide any meaningful result. This is where other techniques, like factorisation and rationalisation, come into play. They help simplify the function or transform it into a form where direct substitution can be applied. It's essential to recognise when direct substitution won't work and to be familiar with alternative methods to evaluate the limit.

Practice Questions

Evaluate the limit of the function f(x) = (x^2 - 4x + 4)/(x - 2) as x approaches 2.

To evaluate this limit, we can use the factorisation method. First, we factorise the numerator:

x2 - 4x + 4 can be written as (x - 2)(x - 2).

Thus, the function becomes:

f(x) = (x - 2)(x - 2)/(x - 2)

Cancelling out the common term x - 2, we get:

f(x) = x - 2

Now, substituting x = 2 into the simplified function, we get:

f(2) = 2 - 2 = 0

Therefore, the limit of f(x) as x approaches 2 is 0.

Evaluate the limit of the function g(x) = (sqrt(x) - 3)/(x - 9) as x approaches 9.

To evaluate this limit, we can use the rationalisation method. We multiply the function by its conjugate:

g(x) = (sqrt(x) - 3)/(x - 9) * (sqrt(x) + 3)/(sqrt(x) + 3)

This simplifies to:

g(x) = (x - 9)/(x - 9) * (sqrt(x) + 3)/(sqrt(x) + 3)

Cancelling out the common term x - 9, we get:

g(x) = 1/ (sqrt(x) + 3)

Now, substituting x = 9 into the simplified function, we get:

g(9) = 1/ (sqrt(9) + 3) = 1/6

Therefore, the limit of g(x) as x approaches 9 is 1/6.

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