Indeterminate Forms

Indeterminate forms are expressions that have an undefined value when evaluated directly. They typically arise when taking limits of functions or evaluating integrals. Some common indeterminate forms include:

• $0/0$ (division by zero)
• $\infty /\infty$ (division of infinity by infinity)
• $0^0$ (zero to the power of zero)
• ${\infty }^0$ (infinity to the power of zero)
• $1^{\infty }$ (one to the power of infinity)
• $\infty -\infty$ (infinity minus infinity)

To evaluate indeterminate forms, we can use various techniques such as L’Hôpital’s rule, series expansions, or algebraic manipulations.

L’Hôpital’s Rule

L’Hôpital’s rule is a powerful technique for evaluating limits of indeterminate forms. It states that if the limit of the numerator and denominator of a fraction is both 0 or both infinity, then the limit of the fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

In other words, if

$$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\frac{0}{0}\text{or}\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\frac{\mathrm{\infty }}{\mathrm{\infty }},$$

then

$$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\underset{x\to a}{lim}\frac{f^{\prime }(x)}{g^{\prime }(x)},$$

provided that the limit on the right side exists.

Examples

Here are some examples of how to evaluate indeterminate forms using L’Hôpital’s rule:

Example 1: Evaluate the limit

$$\underset{x\to 0}{lim}\frac{\sin x}{x}.$$

Solution:

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=\frac{0}{0}.$$

Applying L’Hôpital’s rule, we have

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=\underset{x\to 0}{lim}\frac{\cos x}{1}=1.$$

Example 2: Evaluate the limit

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}.$$

Solution:

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}=\frac{\mathrm{\infty }}{\mathrm{\infty }}.$$

Applying L’Hôpital’s rule, we have

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2}{e^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{2x}{e^x}=\underset{x\to \mathrm{\infty }}{lim}\frac{2}{e^x}=0.$$

Other Techniques

In addition to L’Hôpital’s rule, there are other techniques that can be used to evaluate indeterminate forms. These include:

• Series expansions: Expanding the numerator and denominator of a fraction as power series and then taking the limit term by term.
• Algebraic manipulations: Simplifying the expression by using algebraic identities or factoring.
• Comparison with known limits: Comparing the indeterminate form with known limits of similar expressions.

The choice of technique depends on the specific indeterminate form and the function involved.

Indeterminate Forms List

Indeterminate forms are expressions that have an undefined value when evaluated directly. They typically arise when the limit of a function involves the subtraction or division of two terms that both approach zero or infinity.

The following is a list of common indeterminate forms:

• $0/0$
• $\infty /\infty$
• $0\ast \infty$
• $\infty -\infty$
• $1^{\infty }$
• $0^0$

Evaluating Indeterminate Forms

To evaluate an indeterminate form, we can use various techniques such as L’Hôpital’s rule, series expansions, or algebraic manipulations.

L’Hôpital’s Rule

L’Hôpital’s rule is a powerful technique for evaluating limits of indeterminate forms. It states that if the limit of the numerator and denominator of a fraction both approach zero or infinity, then the limit of the fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

For example, consider the limit:

$$\underset{x\to 0}{lim}\frac{\sin x}{x}$$

Both the numerator and denominator approach zero as $x$ approaches $0$, so we can use L’Hôpital’s rule to evaluate the limit:

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=\underset{x\to 0}{lim}\frac{\cos x}{1}=1$$

Series Expansions

Series expansions can also be used to evaluate indeterminate forms. By expanding the numerator and denominator of a fraction as power series, we can often find the limit of the fraction by taking the limit of the corresponding series.

For example, consider the limit:

$$\underset{x\to 0}{lim}\frac{e^x-1}{x}$$

We can expand $e^x$ as a power series:

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

Substituting this expansion into the fraction, we get:

$$\underset{x\to 0}{lim}\frac{e^x-1}{x}=\underset{x\to 0}{lim}\frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots -1}{x}$$

Simplifying the expression, we get:

$$\underset{x\to 0}{lim}\frac{e^x-1}{x}=\underset{x\to 0}{lim}\frac{x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots }{x}$$

Dividing both the numerator and denominator by $x$, we get:

$$\underset{x\to 0}{lim}\frac{e^x-1}{x}=\underset{x\to 0}{lim}(1+\frac{x}{2!}+\frac{x^2}{3!}+\cdots )$$

Taking the limit, we get:

$$\underset{x\to 0}{lim}\frac{e^x-1}{x}=1$$

Algebraic Manipulations

In some cases, indeterminate forms can be evaluated using algebraic manipulations. For example, consider the limit:

$$\underset{x\to 0}{lim}\frac{x^2-9}{x-3}$$

We can factor the numerator as follows:

$$x^2-9=(x+3)(x-3)$$

Substituting this factorization into the fraction, we get:

$$\underset{x\to 0}{lim}\frac{x^2-9}{x-3}=\underset{x\to 0}{lim}\frac{(x+3)(x-3)}{x-3}$$

Canceling the common factor of $x-3$, we get:

$$\underset{x\to 0}{lim}\frac{x^2-9}{x-3}=\underset{x\to 0}{lim}(x+3)$$

Taking the limit, we get:

$$\underset{x\to 0}{lim}\frac{x^2-9}{x-3}=3$$

Indeterminate form of 0/0

The indeterminate form of 0/0 occurs when both the numerator and denominator of a fraction are zero. This form is indeterminate because it is not immediately clear what the value of the fraction should be. For example, the fraction 0/0 could be equal to 0, 1, or any other number.

Indeterminate Form of $\frac{\mathrm{\infty }}{\mathrm{\infty }}$

In calculus, the indeterminate form $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ occurs when both the numerator and denominator of a fraction approach infinity as the variable approaches a certain value. This form is considered indeterminate because it does not have a definite value. Instead, the limit of the expression may be any real number, including infinity, negative infinity, or a finite value.

To determine the limit of an expression in the form $\frac{\mathrm{\infty }}{\mathrm{\infty }}$, we can use various techniques such as L’Hôpital’s rule, series expansions, or algebraic manipulations.

Indeterminate Form of ∞

In mathematics, an indeterminate form is an expression that has an undefined value. One common type of indeterminate form is 0/0. This occurs when both the numerator and denominator of a fraction are zero. For example, the expression 0/0 is indeterminate because it is not clear what value it should have.

Evaluating Indeterminate Forms

There are a number of techniques that can be used to evaluate indeterminate forms. One common technique is to use L’Hôpital’s rule. This rule states that if the limit of the numerator and denominator of a fraction is both 0 or both infinity, then the limit of the fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

For example, consider the expression 0/0. We can use L’Hôpital’s rule to evaluate this expression by taking the derivative of the numerator and denominator:

$$\underset{x\to 0}{lim}\frac{0}{0}=\underset{x\to 0}{lim}\frac{\frac{d}{dx}(0)}{\frac{d}{dx}(0)}$$

$$=\underset{x\to 0}{lim}\frac{0}{0}$$

Since the limit of the derivative of the numerator and denominator is still 0/0, we need to apply L’Hôpital’s rule again. Taking the second derivative of the numerator and denominator, we get:

$$\underset{x\to 0}{lim}\frac{0}{0}=\underset{x\to 0}{lim}\frac{\frac{d^2}{d{x}^2}(0)}{\frac{d^2}{d{x}^2}(0)}$$

$$=\underset{x\to 0}{lim}\frac{0}{0}$$

Since the limit of the second derivative of the numerator and denominator is still 0/0, we need to apply L’Hôpital’s rule a third time. Taking the third derivative of the numerator and denominator, we get:

$$\underset{x\to 0}{lim}\frac{0}{0}=\underset{x\to 0}{lim}\frac{\frac{d^3}{d{x}^3}(0)}{\frac{d^3}{d{x}^3}(0)}$$

$$=\underset{x\to 0}{lim}\frac{0}{0}$$

Since the limit of the third derivative of the numerator and denominator is still 0/0, we can conclude that the limit of the expression 0/0 is indeterminate.

Other Indeterminate Forms

In addition to 0/0, there are a number of other indeterminate forms. These include:

• $\infty /\infty$
• $0^{\infty }$
• ${\infty }^0$
• $1^{\infty }$
• ${\infty }^1$

Each of these indeterminate forms can be evaluated using a variety of techniques, including L’Hôpital’s rule.

Indeterminate forms are expressions that have an undefined value. There are a number of techniques that can be used to evaluate indeterminate forms, including L’Hôpital’s rule.

Indeterminate Form of $\mathrm{\infty }-\mathrm{\infty }$

In mathematics, the indeterminate form of $\mathrm{\infty }-\mathrm{\infty }$ occurs when the limit of the difference of two divergent sequences or functions is not defined. This means that the result of subtracting one infinity from another infinity can be any real number, including infinity itself, negative infinity, or an undefined value.

Understanding the Indeterminate Form

To understand why $\mathrm{\infty }-\mathrm{\infty }$ is an indeterminate form, consider the following two sequences:

$$a_n=n\text{and}{b}_n=n+1$$

As $n$ approaches infinity, both $a_n$ and $b_n$ approach infinity. However, the difference between these two sequences, $a_n-b_n=-1$, is a constant value that does not approach any specific limit.

This example illustrates that subtracting one infinity from another infinity does not necessarily result in another infinity. In fact, the result can be any real number, depending on the specific sequences or functions involved.

Evaluating Limits with Indeterminate Form

When encountering an indeterminate form of $\mathrm{\infty }-\mathrm{\infty }$, it is necessary to use additional techniques to evaluate the limit. One common method is to factor out the highest power of $n$ from both sequences or functions and then simplify the expression.

For example, consider the following limit:

$$\underset{n\to \mathrm{\infty }}{lim}(n^2-3n+2)-(n^2+2n-1)$$

Factoring out $n^2$ from both terms, we get:

$$\underset{n\to \mathrm{\infty }}{lim}{n}^2(1-\frac{3}{n}+\frac{2}{n^2})-n^2(1+\frac{2}{n}-\frac{1}{n^2})$$

Simplifying the expression, we can cancel out the $n^2$ terms and obtain:

$$\underset{n\to \mathrm{\infty }}{lim}(-4+\frac{5}{n}-\frac{3}{n^2})$$

Taking the limit, we find that the expression converges to $-4$.

The indeterminate form of $\mathrm{\infty }-\mathrm{\infty }$ occurs when the limit of the difference of two divergent sequences or functions is not defined. To evaluate limits with this indeterminate form, it is necessary to use additional techniques such as factoring out the highest power of $n$ and simplifying the expression.

Indeterminate Form of 0${}^0$

The expression $0^0$ is an indeterminate form, meaning that it does not have a definite value. This is because $0^0$ can be equal to any number, depending on the context in which it is used.

Evaluating $0^0$

To evaluate $0^0$, we can use the following steps:

1. Rewrite $0^0$ as $(0{)}^x$, where $x$ is an arbitrary number.
2. Take the limit of $(0{)}^x$ as $x$ approaches 0.

If the limit exists, then it is the value of $0^0$. If the limit does not exist, then $0^0$ is indeterminate.

Examples

Here are some examples of how to evaluate $0^0$:

• $(0{)}^0=1$, because $(0{)}^x$ approaches 1 as $x$ approaches 0.
• $(0{)}^{\infty }=0$, because $(0{)}^x$ approaches 0 as $x$ approaches ∞.
• $(0{)}^{(-\infty )}=\infty$, because $(0{)}^x$ approaches ∞ as $x$ approaches -∞.

Applications

The indeterminate form of $0^0$ has applications in a variety of fields, including mathematics, physics, and engineering. For example, $0^0$ is used to model the behavior of black holes, which are regions of space where the gravitational field is so strong that nothing, not even light, can escape.

The indeterminate form of $0^0$ is a complex and fascinating topic with a wide range of applications. By understanding how to evaluate $0^0$, we can gain a deeper understanding of the world around us.

Indeterminate Form of $1^{\mathrm{\infty }}$

The indeterminate form $1^{\mathrm{\infty }}$ occurs when we have an expression of the form $1^x$ as $x$ approaches infinity. This form is indeterminate because the value of the expression can be either 0 or 1, depending on the specific function involved.

Evaluating $1^{\mathrm{\infty }}$

To evaluate the indeterminate form $1^{\mathrm{\infty }}$, we can use the following steps:

1. Rewrite the expression in terms of logarithms.
2. Apply L’Hôpital’s rule to evaluate the limit of the logarithmic expression.
3. Exponentiate the result to obtain the value of the original expression.

Example

Let’s evaluate the indeterminate form $1^{\mathrm{\infty }}$ for the function $f(x)=(1+1/x{)}^x$.

Step 1: Rewrite the expression in terms of logarithms.

$$\ln f(x)=x\ln (1+1/x)$$

Step 2: Apply L’Hôpital’s rule to evaluate the limit of the logarithmic expression.

$$\underset{x\to \mathrm{\infty }}{lim}x\ln (1+1/x)=\underset{x\to \mathrm{\infty }}{lim}\frac{\ln (1+1/x)}{1/x}$$

$$=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{1+1/x}\cdot (-1/x^2)}{-1/x^2}$$

$$=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{1+1/x}=1$$

Step 3: Exponentiate the result to obtain the value of the original expression.

$$f(x)=e^{\ln f(x)}=e^1=e$$

Therefore, the value of the indeterminate form $1^{\mathrm{\infty }}$ for the function $f(x)=(1+1/x{)}^x$ is $e$.

The indeterminate form $1^{\mathrm{\infty }}$ can be evaluated using L’Hôpital’s rule. By rewriting the expression in terms of logarithms and applying L’Hôpital’s rule, we can determine the value of the expression as $x$ approaches infinity.

Indeterminate Form of ${\mathrm{\infty }}^0$

The indeterminate form ${\mathrm{\infty }}^0$ occurs when the base of an exponential expression approaches infinity and the exponent approaches zero. This form is indeterminate because it can evaluate to any value, including 0, 1, or infinity.

To determine the limit of an expression in the form ${\mathrm{\infty }}^0$, we can use L’Hôpital’s rule. L’Hôpital’s rule states that if the limit of the numerator and denominator of a fraction is both 0 or both infinity, the limit of the fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

Examples

Here are some examples of how to use L’Hôpital’s rule to evaluate limits in the form ${\mathrm{\infty }}^0$:

• $\underset{x\to \mathrm{\infty }}{lim}{x}^0$

Since $x^0=1$ for all values of $x$, the limit of this expression is 1.

• $\underset{x\to 0}{lim}{x}^{\frac{1}{x}}$

Using L’Hôpital’s rule, we have:

$$\underset{x\to 0}{lim}{x}^{\frac{1}{x}}=\underset{x\to 0}{lim}\frac{\ln x}{\frac{1}{x}}=\underset{x\to 0}{lim}\frac{1/x}{-1/x^2}=\underset{x\to 0}{lim}-x=0$$

Therefore, the limit of this expression is 0.

• $\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{x}{)}^x$

Using L’Hôpital’s rule, we have:

$$\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{x}{)}^x=\underset{x\to \mathrm{\infty }}{lim}\frac{\ln (1+\frac{1}{x})}{\frac{1}{x}}=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{1+\frac{1}{x}}\cdot (-\frac{1}{x^2})}{-\frac{1}{x^2}}=\underset{x\to \mathrm{\infty }}{lim}\frac{1}{1+\frac{1}{x}}=1$$

Therefore, the limit of this expression is 1.

The indeterminate form ${\mathrm{\infty }}^0$ can be evaluated using L’Hôpital’s rule. This rule allows us to determine the limit of an expression by taking the derivative of the numerator and denominator and then evaluating the limit of the resulting expression.

Evaluating Indeterminate Form of Limits

When evaluating limits, we sometimes encounter expressions that are indeterminate. This means that the limit does not exist or cannot be determined from the given information. Some common indeterminate forms include:

• $0/0$
• $\infty /\infty$
• $0^0$
• $1^{\infty }$

To evaluate these indeterminate forms, we can use various techniques such as L’Hôpital’s rule, series expansions, or algebraic manipulations.

Series Expansions

Series expansions can also be used to evaluate indeterminate forms. By expanding the numerator and denominator of a fraction as power series, we can often find the limit of the fraction by taking the limit of the corresponding series.

For example, consider the limit

$$\underset{x\to 0}{lim}\frac{\sin x}{x}.$$

We can expand sin(x) as a power series:

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$

Substituting this expansion into the limit, we get

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=\underset{x\to 0}{lim}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots }{x}$$

$$=\underset{x\to 0}{lim}(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots )$$

$$=1.$$

Algebraic Manipulations

In some cases, we can evaluate indeterminate forms using algebraic manipulations. For example, consider the limit

$$\underset{x\to 0}{lim}\frac{x^2-4x+3}{x-1}.$$

We can factor the numerator as follows:

$$x^2-4x+3=(x-1)(x-3).$$

Substituting this factorization into the limit, we get

$$\underset{x\to 0}{lim}\frac{x^2-4x+3}{x-1}=\underset{x\to 0}{lim}\frac{(x-1)(x-3)}{x-1}$$

$$=\underset{x\to 0}{lim}(x-3)$$

$$=-3.$$

Evaluating indeterminate forms of limits can be challenging, but there are various techniques that can be used to find the limit. L’Hôpital’s rule, series expansions, and algebraic manipulations are some of the most commonly used techniques. By understanding these techniques, we can evaluate a wide range of indeterminate forms and determine the limits of functions.

Indeterminate Form Solved Examples

0/0 Form

Example 1:

$$\underset{x\to 0}{lim}\frac{x^2-9}{x-3}$$

Solution:

We can factor the numerator as follows:

$$x^2-9=(x+3)(x-3)$$

Therefore, the limit can be rewritten as:

$$\underset{x\to 0}{lim}\frac{(x+3)(x-3)}{x-3}$$

We can now cancel the common factor of (x - 3) to get:

$$\underset{x\to 0}{lim}(x+3)=3$$

$\infty /\infty$ Form

Example 2:

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2+3x+2}{2x^2-5x+7}$$

Solution:

We can divide both the numerator and the denominator by x^2 to get:

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2+3x+2}{2x^2-5x+7}=\underset{x\to \mathrm{\infty }}{lim}\frac{1+\frac{3}{x}+\frac{2}{x^2}}{2-\frac{5}{x}+\frac{7}{x^2}}$$

As x approaches infinity, all of the terms with x in the denominator will approach zero. Therefore, the limit can be simplified to:

$$\underset{x\to \mathrm{\infty }}{lim}\frac{1+\frac{3}{x}+\frac{2}{x^2}}{2-\frac{5}{x}+\frac{7}{x^2}}=\frac{1}{2}$$

$0\cdot \infty$ Form

Example 3:

$$\underset{x\to 0}{lim}x\ln x$$

Solution:

We can use L’Hôpital’s rule to evaluate this limit. L’Hôpital’s rule states that if the limit of the numerator and denominator of a fraction is both 0 or both infinity, then the limit of the fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

In this case, the derivative of the numerator is ln x and the derivative of the denominator is 1. Therefore, the limit can be rewritten as:

$$\underset{x\to 0}{lim}\frac{x\ln x}{1}=\underset{x\to 0}{lim}\ln x$$

We can now evaluate this limit by taking the natural logarithm of both sides of the equation:

$$\underset{x\to 0}{lim}\ln x=\ln 0=-\mathrm{\infty }$$

Therefore, the limit of the original expression is -∞.

$1^{\infty }$ Form

Example 4:

$$\underset{x\to 0}{lim}(1+x{)}^{1/x}$$

Solution:

We can use the natural logarithm to rewrite this expression as follows:

$$\underset{x\to 0}{lim}(1+x{)}^{1/x}=\underset{x\to 0}{lim}{e}^{\frac{1}{x}\ln (1+x)}$$

We can now use L’Hôpital’s rule to evaluate this limit. L’Hôpital’s rule states that if the limit of the numerator and denominator of a fraction is both 0 or both infinity, then the limit of the fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

In this case, the derivative of the numerator is:

$$\frac{d}{dx}(\frac{1}{x}\ln (1+x))=\frac{1-\ln (1+x)}{x^2}$$

And the derivative of the denominator is:

$$\frac{d}{dx}(1+x)=1$$

Therefore, the limit can be rewritten as:

$$\underset{x\to 0}{lim}\frac{\frac{1-\ln (1+x)}{x^2}}{1}=\underset{x\to 0}{lim}\frac{1-\ln (1+x)}{x^2}$$

We can now evaluate this limit by taking the limit of the numerator and denominator separately. The limit of the numerator is 1, and the limit of the denominator is 0. Therefore, the limit of the original expression is ∞.

Indeterminate Forms FAQs

What is an indeterminate form?

An indeterminate form is a mathematical expression that has an undefined value. This can occur when two expressions approach the same value, but not necessarily the same value. For example, the expression 0/0 is indeterminate because both the numerator and denominator approach zero, but not necessarily at the same rate.

What are the different types of indeterminate forms?

There are three main types of indeterminate forms:

• 0/0: This occurs when both the numerator and denominator of a fraction approach zero.
• ∞/∞: This occurs when both the numerator and denominator of a fraction approach infinity.
• 0^0: This occurs when a number is raised to the power of zero.

How do you evaluate indeterminate forms?

There are a variety of techniques that can be used to evaluate indeterminate forms. Some of the most common techniques include:

• L’Hôpital’s rule: This rule can be used to evaluate indeterminate forms of the type 0/0 and ∞/∞.
• Taylor series: Taylor series can be used to approximate the value of an indeterminate form.
• Limits: Limits can be used to determine the value of an indeterminate form.

What are some examples of indeterminate forms?

Here are some examples of indeterminate forms:

• 0/0: This occurs in expressions such as sin(x)/x as x approaches 0.
• ∞/∞: This occurs in expressions such as x/ln(x) as x approaches infinity.
• 0^0: This occurs in expressions such as 0^x as x approaches 0.

Why are indeterminate forms important?

Indeterminate forms are important because they can lead to incorrect conclusions if they are not handled properly. For example, if you were to evaluate the expression 0/0 without using the proper techniques, you might conclude that the expression is equal to zero. However, this is not necessarily the case. In fact, the expression 0/0 can actually be equal to any number.

Conclusion

Indeterminate forms are a complex topic, but they are important to understand in order to avoid incorrect conclusions. There are a variety of techniques that can be used to evaluate indeterminate forms, and it is important to choose the appropriate technique for the specific form that you are evaluating.