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Chapter 4: Problem 1

Explain with examples what is meant by the indeterminate form \(0 / 0\)

Short Answer

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**Question:** Explain the concept of the indeterminate form \(0 / 0\) and provide examples. **Answer:** The indeterminate form \(0 / 0\) occurs when both the numerator and the denominator of a fraction approach zero. This form is called "indeterminate" because it is impossible to determine its value directly without more context about the function. In some cases, the indeterminate form may result in a finite value, while in other cases, it may result in an infinite value or no value at all. Examples include the limit \(\displaystyle\lim_{x\to 0} \frac{\sin x}{x}\) which resolves to 1, and \(\displaystyle\lim_{x\to 0} \frac{x^2}{x}\) which resolves to 0. However, not all limits with a zero in the denominator will have a finite value, such as \(\displaystyle\lim_{x\to 0} \frac{1}{x^2}\), which does not exist.

Step by step solution

01

Introduction to Indeterminate Form

The indeterminate form \(0 / 0\) occurs when both the numerator and the denominator of a fraction approach zero. This form is called "indeterminate" because it is impossible to determine its value directly without more context about the function. In some cases, the indeterminate form may result in a finite value, while in other cases, it may result in an infinite value or no value at all.
02

Example 1: Evaluating a limit with indeterminate form

Consider the function \(\displaystyle\lim_{x\to 0} \frac{\sin x}{x}\). Directly plugging in \(x=0\) in the function, we get \(\frac{\sin 0}{0} = \frac{0}{0}\), which is in indeterminate form. However, using L'H么pital's rule, where we take the derivative of the numerator and denominator, we get \(\displaystyle\lim_{x\to 0} \frac{\cos x}{1} = \cos 0 = 1\). So, in this case, the indeterminate form \(0 / 0\) results in a finite value, 1.
03

Example 2: Another limit with indeterminate form

Let's look at another example, \(\displaystyle\lim_{x\to 0} \frac{x^2}{x}\). Plugging in \(x=0\), we get \(\frac{0^2}{0} = \frac{0}{0}\), again an indeterminate form. Taking the limit using algebraic simplification, we can cancel the common factor x in the numerator and denominator: \(\displaystyle\lim_{x\to 0} x\). When we plug in \(x=0\), we get \(0\). In this example, the indeterminate form \(0 / 0\) also results in a finite value, 0.
04

Example 3: A limit with no finite result

Now let's consider \(\displaystyle\lim_{x\to 0} \frac{1}{x^2}\). When we plug in \(x=0\), we get \(\displaystyle\frac{1}{0^2} = \frac{1}{0}\). This is not of the form \(0 / 0\), but it shows that not all limits with a zero in the denominator will have a finite value. In this case, the limit does not exist because the function approaches infinity as \(x\) approaches zero. In summary, the indeterminate form \(0 / 0\) highlights a situation where we cannot directly determine the value of a function without additional information, as is the case when evaluating limits. Understanding this concept is crucial in calculus, as it emphasizes the importance of using proper techniques and rules for working with limits and derivatives.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limits in Calculus
In calculus, limits are fundamental to the study of functions, their continuity, and rates of change. The limit of a function at a given point describes the value that the function approaches as the input (or 'x' value) gets closer to that point. Limits help in understanding the behavior of functions, especially in scenarios where the function may not be well-defined or easily evaluated, such as the indeterminate form 0/0.

For example, if we encounter a situation where substituting a number into a function provides an undefined result, such as dividing by zero, we must resort to evaluating the limit to comprehend the function's behavior around that point. Understanding limits is critical to progressing in calculus, as they form the basis for defining derivatives and integrals.
L'H么pital's Rule
When faced with an indeterminate form like 0/0, L'H么pital's rule can often come to the rescue. This powerful tool in calculus is used for evaluating the limits of functions that yield an indeterminate form. According to L'H么pital's rule, if the limits of both the numerator and denominator of a fraction are zero or both are infinite, the limit of the fraction may be found by taking the derivative of the numerator and the derivative of the denominator and then finding the limit of the new fraction.

L'H么pital's rule transforms an indeterminate form into a more workable expression whose limit can be evaluated directly or further simplified. This rule is exceptionally useful when dealing with complicated limits and has the potential to simplify the process of finding the limit significantly.
Evaluating Limits
Evaluating limits often requires a combination of techniques depending on the form of the function and the point at which the limit is being evaluated. When encountering the indeterminate form 0/0, it's important to recognize that the limit can exist even if the function itself is not defined at that point. Techniques to evaluate these limits include factoring, rationalizing expressions, or using L'H么pital's rule if the conditions permit.

Understanding how to evaluate these limits ensures that one can accurately determine function behavior at points that are not immediately clear. Moreover, it provides a foundation for exploring other complex topics in calculus such as optimization and integration.
Algebraic Simplification
Algebraic simplification is a valuable technique in evaluating limits, especially when dealing with indeterminate forms like 0/0. Simplification can involve canceling out common factors, expanding expressions, or manipulating the function algebraically to eliminate the indeterminate nature of the limit. It's often the first method attempted before considering more advanced techniques like L'H么pital's rule.

For instance, when a term in both the numerator and the denominator approaches zero, sometimes they can be factored and reduced. This approach is not only simpler but also highlights the importance of having strong algebraic skills when studying calculus. Simplification can turn a daunting expression into a tractable problem.

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Most popular questions from this chapter

Find the solution of the following initial value problems. \(y^{\prime}(\theta)=\frac{\sqrt{2} \cos ^{3} \theta+1}{\cos ^{2} \theta} ; y\left(\frac{\pi}{4}\right)=3,-\pi / 2<\theta<\pi / 2\)

Determine the following indefinite integrals. Check your work by differentiation. $$\int \frac{v^{3}+v+1}{1+v^{2}} d v$$

Use limit methods to determine which of the two given functions grows faster, or state that they have comparable growth rates. $$x^{2} \ln x ; \ln ^{2} x$$

Exponentials vs. superexponentials Show that \(x^{x}\) grows faster than \(b^{x}\) as \(x \rightarrow \infty,\) for \(b>1\)

Use limit methods to determine which of the two given functions grows faster, or state that they have comparable growth rates. $$e^{x^{2}} ; x^{x / 10}$$
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