Compute limits using algebraic techniques.
Limit laws
In section 1.2, Numerical Limits, we briefly discussed “plugging in” to compute limits. The example we investigated was
The reason plugging in worked in the previous example is a direct consequence of the limit laws presented below.
In each of the following laws, all of the limits are assumed to exist.
- 1.
- The limit of a constant is the constant:
- 2.
- The next law is self-evident:
- 3.
- The limit of a multiple of a function is the multiple of the limit:
- 4.
- The limit of a sum is the sum of the limits:
- 5.
- The limit of a difference is the difference of the limits:
- 6.
- The limit of a product is the product of the limits:
- 7.
- The limit of a quotient is the quotient of the limits: provided that the limit in the denominator is not equal to zero.
- 8.
- Limits can be moved inside of radicals:
Each of the above limit laws is valid if or .
Factor and cancel
example 2 Compute the limit:
Plugging in the terminal value, , yields the indeterminate form . To find this limit analytically, we will factor the numerator and denominator and simplify the fraction. In the numerator we have a difference of squares, and such a form always factors in the following way: Applying this general formula to our example, we get Next, in the denominator, we can factor out a common factor of from each of the terms: With these factorizations, we can simplify the fraction and find the limit:
Hence, This example uses the factor and cancel method.
example 3 Compute the limit:
Plugging in the terminal value yields the indeterminate form . The numerator factors by finding two numbers which multiply to give and add to give . The two numbers are and , hence the factorization is In the denominator, we have a difference of cubes, and we use the formula: Applying this to our example gives With these factorizations, we can simplify the fraction and find the limit:
Plugging in the terminal value yields the indeterminate form . The numerator factors by finding two numbers which multiply to give and add to give . The two numbers are and , hence the factorization is In the denominator, we have a difference of cubes, and we use the formula: Applying this to our example gives With these factorizations, we can simplify the fraction and find the limit:
Hence,
(problem 3a) Compute the following limit which we investigated numerically in the
previous section:
The value of the limit is
example 4 Compute the limit:
Plugging in the terminal value yields the indeterminate form . To factor the numerator and denominator in this case, we will use the important fact that for a polynomial and a number , And once we know one factor of a polynomial, it is usually much easier to find the other. Since plugging in gave in both the numerator and the denominator, both polynomials have as a factor. In the numerator, the factorization is and in the denominator, the factorization is .
Plugging in the terminal value yields the indeterminate form . To factor the numerator and denominator in this case, we will use the important fact that for a polynomial and a number , And once we know one factor of a polynomial, it is usually much easier to find the other. Since plugging in gave in both the numerator and the denominator, both polynomials have as a factor. In the numerator, the factorization is and in the denominator, the factorization is .
With these factorizations, we can simplify the fraction and find the limit:
Hence,
Conjugate radicals
The expressions are called conjugate radicals. When we multiply conjugate radicals using the difference of squares formula, , we get an expression that is free of radicals:
We will now take advantage of this to find limits.
example 5 Compute the limit:
To solve this limit problem, we will use the conjugate radical of the numerator, which is . In order to maintain the value of the expression given in the problem, we multiply by one in the form of the conjugate radical over itself:
In the last step, we plugged in the terminal value to get the final answer .
example 6 Compute the limit:
To solve this limit problem, we will use the conjugate radical of the numerator, which is . In order to not change the value of the expression given in the problem, we multiply by one in the form of the conjugate radical divided by itself:
In the last step, we plugged in the terminal value to get the final answer .
example 7 Compute the limit: .
If we plug in , the numerator and denominator are both 0. The conjugate of the radical expression in the denominator is To simplify our calculations a little bit, let’s multiply the conjugates together separately: With this in mind, we have:
If we plug in , the numerator and denominator are both 0. The conjugate of the radical expression in the denominator is To simplify our calculations a little bit, let’s multiply the conjugates together separately: With this in mind, we have:
Complex fractions
We now consider examples involving fractions within fractions, called complex fractions.
example 8 Compute the limit:
Plugging in yields the familiar indeterminate form. The algebra skills necessary to transform the function in the problem to one which allows plugging in involve subtraction and division of fractions. The rules are summarized as: Applying these to our problem, we get
Plugging in yields the familiar indeterminate form. The algebra skills necessary to transform the function in the problem to one which allows plugging in involve subtraction and division of fractions. The rules are summarized as: Applying these to our problem, we get
It is important to note that and are opposites which cancel, leaving : The ratio of opposites is .
example 9 Compute the limit: Carefully plugging in gives , so we
begin simplifying the complex fraction. Since this function is bulky,
let’s do the addition of fractions from the numerator separately:
Now back to the original problem:
Absolute values
The definition of the absolute value is:
To calculate a limit involving an absolute value, we will need to remove the absolute value bars. To do this correctly, we can see from the definition that it is necessary to know whether the quantity in the absolute value bars is positive or negative.
example 10 Compute the limit: .
Plugging in gives the indeterminate form . To resolve this limit, we will do a sign analysis on the quantity in the absolute value bars, .
Plugging in gives the indeterminate form . To resolve this limit, we will do a sign analysis on the quantity in the absolute value bars, .
Since , we have and hence . Since the quantity in the absolute value bars is negative, we can compute its absolute value as follows:
Using this in the limit, we get
Here are some detailed, lecture style videos on finding limits analytically:
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