/*! 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 36 Gives a function \(f(x),\) a poi... [FREE SOLUTION] | 91Ó°ÊÓ

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Gives a function \(f(x),\) a point \(c,\) and a positive number \(\epsilon .\) Find \(L=\lim _{x \rightarrow c} f(x) .\) Then find a number \(\delta>0\) such that for all \(x\) \(0<|x-c|<\delta \quad \Rightarrow \quad|f(x)-L|<\epsilon\). \(f(x)=4 / x, \quad c=2, \quad \epsilon=0.4\)

Short Answer

Expert verified
\(\lim_{x \to 2} \frac{4}{x} = 2\), and \(\delta = 0.33\).

Step by step solution

01

Calculate the Limit

First, we need to find the limit \(L\) of the function \(f(x) = \frac{4}{x}\) as \(x\) approaches \(c = 2\). The limit can be calculated as follows:\[L = \lim_{x \to 2} \frac{4}{x} = \frac{4}{2} = 2.\]
02

Define the Epsilon-Delta Condition

We want to find a \(\delta > 0\) such that whenever \(0 < |x - 2| < \delta\), it follows that \(|f(x) - L| < \epsilon\). Here, \(\epsilon = 0.4\) and \(L = 2\). Thus, we set up the inequality:\[|\frac{4}{x} - 2| < 0.4.\]
03

Solve the Inequality for Delta

Rewriting the inequality \(|\frac{4}{x} - 2| < 0.4\) involves solving:\[|\frac{4 - 2x}{x}| < 0.4.\]Split the inequality:\[-0.4 < \frac{4 - 2x}{x} < 0.4.\]
04

Solve the First Part of the Inequality

Solve \(\frac{4 - 2x}{x} > -0.4\) by multiplying through by \(x\) (assuming \(x > 0\)):\[4 - 2x > -0.4x.\]Combine similar terms and solve for \(x\):\[4 > 1.6x \Rightarrow x < \frac{4}{1.6} = 2.5.\]
05

Solve the Second Part of the Inequality

For \(\frac{4 - 2x}{x} < 0.4\), multiply through by \(x\) to obtain:\[4 - 2x < 0.4x.\]Combine similar terms and solve for \(x\):\[4 < 2.4x \Rightarrow x > \frac{4}{2.4} \approx 1.67.\]
06

Combine the Results

From Steps 4 and 5, we find \(1.67 < x < 2.5\). The maximum \(\delta\) that satisfies this interval and the condition \(0 < |x - 2| < \delta\) is the smallest distance from 2 to either bounds of the interval, which is:\[\delta = \min(2 - 1.67, 2.5 - 2) = \min(0.33, 0.5) = 0.33.\]

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Key Concepts

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

Understanding Limits
Limits are a foundational concept in calculus and are used to describe the behavior of functions as they approach a certain point. When we say that the limit of a function \( f(x) \) as \( x \) approaches \( c \) is \( L \), it means that as \( x \) gets closer and closer to \( c \), \( f(x) \) also gets closer to \( L \).
This idea is like predicting where a moving object will end up based on its current direction and speed.Key points about limits:
  • They help understand the behavior of functions near a specific point even if the function is not defined exactly at that point.
  • They provide vital information about function trends and patterns.
  • Formally, limits are written as \( \lim_{x \to c} f(x) = L \).
In our original exercise, we calculated the limit of \( f(x) = \frac{4}{x} \) as \( x \to 2 \). By simple substitution, we found \( \lim_{x \to 2} \frac{4}{x} = 2 \). This tells us that as \( x \) approaches 2, \( f(x) \) approaches 2.
Continuity in a Function
Continuity is closely related to limits. A function is continuous at a point \( c \) if the limit of the function as \( x \) approaches \( c \) is equal to the function's value at \( c \). This means there are no jumps, breaks, or holes in the graph at that point. It's like drawing a line without lifting your pencil off the paper.For a function \( f(x) \) to be continuous at a point \( c \), three conditions must be met:
  • \( f(c) \) must be defined.
  • The limit \( \lim_{x \to c} f(x) \) must exist.
  • \( \lim_{x \to c} f(x) = f(c) \)
In our context, since \( \lim_{x \to 2} \frac{4}{x} = 2 \) and \( f(2) = 2 \), the function is continuous at that point. Understanding continuity helps us ensure functions behave predictably near specific points.
Core Calculus Concepts
Calculus is built around a few fundamental ideas, notably limits, derivatives, and integrals. These concepts allow us to explore changes and patterns and to solve complex, real-world problems.
The epsilon-delta definition of a limit is a formal way to capture the notion of a limit and is crucial for addressing mathematical strictness and accuracy.Here’s how epsilon-delta comes into play:
  • \( \epsilon \) (epsilon) represents how close you want \( f(x) \) to be near \( L \).
  • \( \delta \) (delta) represents how close \( x \) needs to be to \( c \) for \( f(x) \) to be within \( \epsilon \) of \( L \).
  • The challenge is to find a \( \delta \) such that whenever \( 0 < |x - c| < \delta \), it follows that \( |f(x) - L| < \epsilon \).
This precise definition allows mathematicians and students alike to have a deeper understanding of how limits and continuous functions work, ensuring problem-solving in calculus is built on solid ground. Our exercise showed this by solving inequalities to find a possible \( \delta = 0.33 \) for our given function and parameters.

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