/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 A function \(f\) and a point \(c... [FREE SOLUTION] | 91Ó°ÊÓ

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A function \(f\) and a point \(c\) not in the domain of \(f\) are given. Analyze \(\lim _{x \rightarrow c} f(x)\) as follows. a. Evaluate \(f\left(c-1 / 10^{n}\right)\) and \(f\left(c+1 / 10^{n}\right)\) for \(n=2,3,4\). b. Formulate a guess for the value \(\ell=\lim _{x \rightarrow c} f(x)\). c. Find a value \(\delta\) such that \(f(x)\) is within 0.01 of \(\ell\) for every \(x\) that is within \(\delta\) of \(c\). d. Graph \(y=f(x)\) for \(x\) in \(\left[c-1 / 10^{4}, c\right) \cup\left(c, c+1 / 10^{4}\right]\) to verify visually that the limit of \(f\) at \(c\) exists. $$ f(x)=\sin (x) / x, c=0 $$

Short Answer

Expert verified
The limit \(\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1\), with \(\delta = 0.1\) keeping \(f(x)\) within 0.01 of 1.

Step by step solution

01

Evaluate Function Values (Left)

Evaluate the function at points approaching from the left, that is, at \(f\left(c-1 / 10^n\right)\) for \(n = 2, 3, 4\). Since \(c = 0\), we have:\[f\left(-1/10^2\right) = f\left(-0.01\right) = \frac{\sin(-0.01)}{-0.01} \approx 0.999983\]\[f\left(-1/10^3\right) = f\left(-0.001\right) = \frac{\sin(-0.001)}{-0.001} \approx 0.9999998\]\[f\left(-1/10^4\right) = f\left(-0.0001\right) = \frac{\sin(-0.0001)}{-0.0001} \approx 0.999999998\]
02

Evaluate Function Values (Right)

Evaluate the function at points approaching from the right, that is, at \(f\left(c+1 / 10^n\right)\) for \(n = 2, 3, 4\). Since \(c = 0\), we have:\[f\left(1/10^2\right) = f\left(0.01\right) = \frac{\sin(0.01)}{0.01} \approx 0.999983\]\[f\left(1/10^3\right) = f\left(0.001\right) = \frac{\sin(0.001)}{0.001} \approx 0.9999998\]\[f\left(1/10^4\right) = f\left(0.0001\right) = \frac{\sin(0.0001)}{0.0001} \approx 0.999999998\]
03

Formulate a Limit Guess

From the calculations in Steps 1 and 2, observe that as \(n\) increases, the values of \(f\left(c \pm 1/10^n\right)\) approach 1. Therefore, the guess for the limit is \(\ell = \lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1\).
04

Determine Delta for Limit Condition

To find \(\delta\) such that \(f(x)\) is within 0.01 of \(\ell = 1\) for every \(x\) within \(\delta\) of \(c\), we need \(-0.01 < f(x) - 1 < 0.01\). Numerically, calculate \(\left|\frac{\sin(x)}{x} - 1\right| < 0.01\), which occurs when \(|x| < \delta = 0.1\).
05

Verify with Graph

Graph the function \(y = \frac{\sin(x)}{x}\) for \(x\) in the interval \([-0.0001, 0) \cup (0, 0.0001]\). Visually confirm that as \(x\) approaches 0 from both directions, the function value urges towards 1, supporting the limit \(\ell = 1\).

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Key Concepts

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

Function Analysis
When analyzing a function, our goal is to understand its behavior at different points, including those that are not in its domain. In this exercise, the function we are dealing with is \( f(x) = \frac{\sin(x)}{x} \), with a focus on the point \( c = 0 \), which is outside the function's domain since dividing by zero is undefined. Analyzing a function involves evaluating its limits, which helps us determine how the function behaves as it approaches certain values.
To start this analysis, compare the values of the function as \( x \) approaches zero from both the left (negative) and the right (positive) sides. This method of function analysis, through asymmetric limits, helps clarify the behavior and tendencies of the function near the undefined point.
Sine Function
The sine function, represented as \( \sin(x) \), is a periodic and oscillating function crucial in trigonometry and calculus. In our case, we are examining the expression \( \frac{\sin(x)}{x} \), a well-known limit problem in calculus.
The sine function has interesting properties, such as repeating every \(2\pi\) radians and being an odd function, meaning \( \sin(-x) = -\sin(x) \). These properties justify certain behaviors of the function as \( x \) approaches zero. As \( x \) gets smaller and approaches zero, the value of \( \sin(x) \) is approximately equal to \( x \) itself, which is an essential aspect contributing to the limit of \( \frac{\sin(x)}{x} \) approaching 1.
Limit Guessing
Limit guessing involves approximating the value that a function approaches as the input value gets close to a point. In this exercise, we are trying to find the limit \( \ell = \lim_{x \to 0} \frac{\sin(x)}{x} \). By observing function values as they approach from both the left and the right, we can form a hypothesis.
When computing the function for small values near the problematic point \( c = 0 \), both from negative and positive sides, we notice a pattern: the values are very close to 1. This suggests that as \( x \rightarrow 0 \), the function values are converging towards 1, leading us to guess that \( \ell = 1 \). Thus, the limit guessing approach combines numerical exploration and pattern recognition to arrive at a plausible limit prediction.
Delta-Epsilon Definition
The delta-epsilon definition is a formal method for proving the limits of a function. To say \( \lim_{x \to c} f(x) = \ell \) means that for every epsilon \( \varepsilon > 0 \), there exists a delta \( \delta > 0 \) such that if \( 0 < |x - c| < \delta \), then \( |f(x) - \ell| < \varepsilon \).
In practical terms, this means that we can make \( f(x) \) as close to \( \ell \) as we desire by selecting \( x \) sufficiently close to \( c \). For our problem, we determined that choosing \( \delta = 0.1 \) ensures \( |\frac{\sin(x)}{x} - 1| < 0.01 \) for \( x \) near zero. This confirms the limit \( \ell = 1 \) with precise and controlled accuracy, fully supporting our earlier guess and specifying with mathematical rigor the limit behavior of the function.

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Most popular questions from this chapter

In each of Exercises \(93-96,\) plot the given functions \(g, f,\) and \(h\) in a common viewing rectangle that illustrates \(f\) being pinched at the point \(c\). Determine \(\lim _{x \rightarrow c} f(x)\). $$ \begin{array}{l} g(x)=(x-2)^{2}, f(x)=(x-2)^{2}(2+\sin (\tan (\pi / x))) \\\ (x-2)^{2}, h(x)=3(x-2)^{2} \quad c=2 \end{array} $$

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Prove that a function \(f\) is continuous at \(c\) if and only if \(f\) is both left-continuous at \(c\) and right-continuous at \(c\).

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