/*! 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 43 Find the limit \(L\). Then use t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the limit \(L\). Then use the \(\varepsilon-\delta\) definition to prove that the limit is \(L\). $$ \lim _{x \rightarrow 0} \sqrt[3]{x} $$

Short Answer

Expert verified
The limit \(L\) of the function \(\sqrt[3]{x}\) as \(x\) approaches 0 is 0. The limit can be proved by the \(\varepsilon-\delta\) definition by setting \(\delta = \varepsilon^3\).

Step by step solution

01

Compute the limit

First, we need to compute the limit of the function as \(x\) approaches 0. For \(f(x) = \sqrt[3]{x}\), it's clear that if \(x\) approaches 0, \(f(x)\) will also approach 0 since \(\sqrt[3]{0}=0\). Therefore, \(L = 0\)\.
02

Understand the \(\varepsilon-\delta\) definition

Next, one must grasp the \(\varepsilon-\delta\) definition of a limit. According this definition, for any number \(\varepsilon > 0\), no matter how small, there is a corresponding number \(\delta > 0\) such that if the distance of \(x\) from 0 (the point of convergence) is less than \(\delta\) (but not equal to 0), then the distance of \(\sqrt[3]{x}\) from \(L=0\) will be less than \(\varepsilon\). In simpler terms, this means if \(0 < |x-0| < \delta\) then \(|\sqrt[3]{x} - 0| < \varepsilon\). It can be deduced from the original inequality \(|\sqrt[3]{x} - 0| < \varepsilon\) that \(|x| < \varepsilon^3\). Hence, we can select \(\delta\) as \(\varepsilon^3\).\.
03

Proof

Since we have found a suitable \(\delta\) for any \(\varepsilon > 0\), we've fulfilled the \(\varepsilon-\delta\) definition. Consider any \(\varepsilon > 0\). If \(0 < |x-0| < \delta\) (or \(0 < |x| < \delta\)), then \(|\sqrt[3]{x} - 0| = |\sqrt[3]{x}| = \sqrt[3]{|x|} < \sqrt[3]{\delta} = \sqrt[3]{\varepsilon^3} = \varepsilon\). Hence, the limit \(L=0\) is proved by the \(\varepsilon-\delta\) definition.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Epsilon-Delta Definition
The epsilon-delta ( \( \varepsilon-\delta \) ) definition is a formal method used to prove the limit of a function. It provides a precise way to say that a function approaches a particular value as the input approaches a certain point.
To understand this concept, it's helpful to simplify the idea. The definition says: for every number \( \varepsilon > 0 \), no matter how tiny, there exists another number \( \delta > 0 \), such that whenever \(|x - a| < \delta \), the function's value \(|f(x) - L| < \varepsilon \). Here, \( a \) is the point that \( x \) is approaching, and \( L \) is the limit.
  • It's like saying: If you want the function's output to be super close to \( L \), then you can always find an interval around \( a \) where the function behaves the way you want it to.
  • The choice of \( \delta \) ensures the function remains close to the desired limit \( L \), as required by believing in \( |f(x) - L| < \varepsilon \).
This framework is what gives limits their mathematical rigor, ensuring they work as expected for any level of precision.
Cubic Root Function
A cubic root function is a type of radical function that extracts the cube root of any real number. The notation is \( \sqrt[3]{x} \), which represents finding a number \( y \) such that \( y^3 = x \). This function is defined for all real numbers as every real number has a cube root.
The cube root has a unique property in that it works both for negative and positive numbers since cubing a negative number results in a negative value, and cubing a positive number results in a positive value.
  • This symmetry implies \( \sqrt[3]{x} \) behaves predictably across the entire number line.
  • When thinking about limits at a point like 0, it's essential to realize that as you approach \( x \rightarrow 0 \), the function output \( \sqrt[3]{x} \rightarrow 0 \), illustrating continuity and the properties of the cubic root at this point.
Cubic functions are less commonly discussed than square roots,
but they offer a great deal of insight into symmetry and the behavior of roots for any given real number.
Limit Proof
In mathematical terms, proving a limit means demonstrating that for every small number \( \varepsilon > 0 \), there is a \( \delta > 0 \) such that whenever \( |x - 0| < \delta \), the condition \( |\sqrt[3]{x} - 0| < \varepsilon \) is satisfied.
We break this down into steps for better understanding:
  • Recognize the specific limit problem. Here, we want to show that \( \lim_{x \to 0} \sqrt[3]{x} = 0 \).
  • The goal is to find a \( \delta \) that assures us that \( |\sqrt[3]{x}| < \varepsilon \). For the cubic root function, this leads us to recognize that \( |x| < \varepsilon^3 \), giving our potential \( \delta \).
Hence, by setting \( \delta = \varepsilon^3 \), we confirm that for all valid \( \varepsilon \), there indeed exists the required \( \delta \) demonstrating the limit holds true. This concrete case exemplifies how the epsilon-delta approach finds specific conditions to ensure desired outcomes, proving the limits effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.