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Chapter 2: Problem 4

\(\lim _{x \rightarrow 0} \frac{x}{x}\) is (A) 1 (B) 0 (C) -1 (D) nonexistent

Short Answer

Expert verified
The limit is 1 (answer A).

Step by step solution

01

Identify the Function

The given function is \( f(x) = \frac{x}{x} \) for \( x eq 0 \). Hence, for all \( x eq 0 \), \( \frac{x}{x} = 1 \).
02

Consider the Limit Approach

The limit expression \( \lim_{{x \to 0}} \frac{x}{x} \) should be evaluated by considering the values of \( x \) as it approaches 0, excluding 0 itself. Therefore, \( f(x) = 1 \) for all values around 0.
03

Evaluate the Limit

Since \( f(x) = 1 \) for all \( x eq 0 \), the limit \( \lim_{{x \to 0}} \frac{x}{x} = \lim_{{x \to 0}} 1 = 1 \). Regardless of whether \( x \) approaches from the positive or negative side, the function remains \( 1 \).
04

Conclude the Limit Existence

The function consistently equals 1 in the neighborhood of 0 (excluding 0 itself). Therefore, the limit is 1, and it exists.

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Key Concepts

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

calculus
Calculus is a branch of mathematics that deals with change and motion, extending geometry and algebra. It focuses on two main concepts: differentiation and integration. Differentiation is about finding the rate at which things change, while integration is about accumulating values.
  • Calculus allows us to model real-world situations mathematically.
  • It provides tools to analyze curves, motion, area, and volume.
In the context of limits, calculus provides a way to examine how a function behaves as its input gets close to a particular point, often dealing with indeterminate forms where functions are not directly evaluable. These aspects make calculus essential for understanding complex systems in both theoretical and applied sciences.
limit evaluation
Limit evaluation helps us determine the value a function approaches as its variable gets closer to a specific point. Limits are foundational for calculus and are often expressed as \( \lim_{{x \to a}} f(x) \).
  • Limits provide a precise way to discuss function behavior.
  • They are essential in defining concepts like continuity and the derivative.
When evaluating limits, it's important to focus on the behavior of the function as it nears the point of interest, without necessarily reaching it. In the exercise, \( \lim_{{x \to 0}} \frac{x}{x} \) is evaluated by considering that for all \( x eq 0 \), \( \frac{x}{x} = 1 \). This understanding relies on recognizing that the direct substitution of \( x = 0 \) results in an indeterminate form.
function behavior near point
Understanding function behavior near a particular point involves analyzing how a function acts as the input values approach that point. This concept is critical in limit evaluation, as it helps determine whether a limit exists, and if so, what its value is.
  • Function behavior can vary depending on which direction values approach from.
  • Analyzing this behavior helps in understanding continuity and potential discontinuities.
In the case of \( \frac{x}{x} \), except at \( x = 0 \), the function equals 1, indicating that as \( x \) approaches 0, the function consistently behaves like 1. This consistent behavior around, but not at, the point leads us to conclude that the limit is 1. Recognizing this requires an understanding of one-sided limits, examining the approach from both positive and negative directions.
indeterminate forms
Indeterminate forms arise in calculus when an expression takes on a form where direct evaluation does not easily reveal the behavior of a function. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), etc.
  • These forms signal the need for deeper analysis through limit evaluation or algebraic manipulation.
  • Understanding them is crucial for accurately interpreting the behavior of functions near challenging points.
With \( \frac{x}{x} \) as \( x \to 0 \), we encounter \( \frac{0}{0} \), prompting us to explore the expression excluding the zero directly. Since for every point except where \( x = 0 \), \( \frac{x}{x} = 1 \), the limit exists and is 1, avoiding the indeterminate form entirely. Addressing indeterminate forms often involves applying L'H么pital's Rule or algebraic simplifications to find a defined limit.

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Most popular questions from this chapter

The graph of \(f(x)=\frac{4}{x^{2}-1}\) has (A) one vertical asymptote, at \(x=1\) (B) the \(y\) -axis as its vertical asymptote (C) the \(x\) -axis as its horizontal asymptote and \(x=\pm 1\) as its vertical asymptotes (D) two vertical asymptotes, at \(x=\pm 1,\) but no horizontal asymptote

\(\lim _{x \rightarrow \infty} \frac{2^{-x}}{2^{2}}\) is (A) -1 (B) 1 (C) 0 (D) \(\infty\)

Which statement is true about the curve \(y=\frac{2 x^{2}+4}{2+7 x-4 x^{2}} ?\) (A) The line \(x=-\frac{1}{4}\) is a vertical asymptote. (B) The line \(x=1\) is a vertical asymptote. (C) The line \(y=-\frac{1}{4}\) is a horizontal asymptote. (D) The line \(y=2\) is a horizontal asymptote.

\(\lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{(\pi-x)}\) is (A) 1 (B) 0 (C) \(\pi\) (D) nonexistent

\(\lim _{x \rightarrow \infty} \frac{2 x^{2}+1}{(2-x)(2+x)}\) is (A) -2 (B) (C) 2 (D) nonexistent
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