Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{1-e^{x^{3}}}{x^{3}}$$

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Step by step solution

01

Recall the Maclaurin series for \(e^x\)

The Maclaurin series for \(e^x\) is given by: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \] Our goal is to find the series for \(e^{x^3}\).
02

Substitute \(x^3\) into the Maclaurin series for \(e^x\)

Substitute \(x^3\) in place of \(x\) in the series: \[ e^{x^3} = 1 + x^3 + \frac{{(x^3)}^2}{2!} + \frac{{(x^3)}^3}{3!} + \cdots \] Simplify the terms: \[ e^{x^3} = 1 + x^3 + \frac{x^6}{2!} + \frac{x^9}{3!} + \cdots \]
03

Expand \(1 - e^{x^3}\)

Subtract the series we found from 1: \[ 1 - e^{x^3} = 1 - \left( 1 + x^3 + \frac{x^6}{2!} + \frac{x^9}{3!} + \cdots \right) = -x^3 - \frac{x^6}{2!} - \frac{x^9}{3!} - \cdots \]
04

Formulate the given limit expression

Substitute the expanded series into the limit: \[ \lim_{x \to 0} \frac{1 - e^{x^3}}{x^3} = \lim_{x \to 0} \frac{-x^3 - \frac{x^6}{2!} - \frac{x^9}{3!} - \cdots}{x^3} \]
05

Simplify the limit

Divide each term in the numerator by \(x^3\): \[ \lim_{x \to 0} \left( \frac{-x^3}{x^3} + \frac{-x^6}{2! x^3} + \frac{-x^9}{3! x^3} + \cdots \right) = \lim_{x \to 0} \left( -1 - \frac{x^3}{2!} - \frac{x^6}{3!} - \cdots \right) \]
06

Evaluate the limit as \(x\) approaches 0

As \(x\) approaches 0, all higher-order terms involving \(x\) will tend to 0. Thus, the limit evaluates to: \[ \lim_{x \to 0} \left( -1 - 0 - 0 - \cdots \right) = -1 \]

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Key Concepts

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

limit evaluation
Understanding limit evaluation is crucial in calculus. Limits help us determine the behavior of functions as they approach a specific point. In this exercise, we used the limit as \(x\) approaches 0 for the function \( \frac{1-e^{x^{3}}}{x^{3}} \). By simplifying the expression using the Maclaurin series expansion, we managed to break down the complex exponential function into simpler terms. This allowed us to observe the behavior of each term as \(x\) neared 0. By the end, higher-order terms became negligible, leaving us with the simplified limit. This shows how limits can reduce complex equations to simpler forms, making them easier to analyze.
series expansion
Series expansion is a powerful tool in calculus. It allows us to write a function as an infinite sum of its terms. The Maclaurin series is a specific type of series expansion centered around 0. For an exponential function like \(e^x\), its Maclaurin series is:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
This series helps to approximate functions and perform complex calculations quickly. In the given problem, we expanded \(e^{x^3}\) using the Maclaurin series. By substituting \(x^3\) into the series, we obtained:
\[ e^{x^3} = 1 + x^3 + \frac{x^6}{2!} + \frac{x^9}{3!} + \cdots \]
This step allowed us to simplify the original limit problem, demonstrating the utility of series expansions in simplifying complex variables.
exponential functions
Exponential functions, denoted as \(e^x\), are fundamental in calculus and many applied fields. They exhibit continuous growth and are central in modeling real-world phenomena, such as population growth and radioactive decay.
A key property of exponential functions is their derivative, which is unique because the derivative of \(e^x\) is also \(e^x\). In this exercise, we dealt with the exponential function \(e^{x^3}\) through its series expansion.
By expanding \(e^{x^3}\) into the series, we were able to simplify the given problem. We transformed a complex exponential term into manageable polynomial terms, facilitating the limit evaluation process.
calculus
Calculus is the mathematical study of change, and it consists of two major branches: differential calculus and integral calculus. Differential calculus focuses on the concept of the derivative, which represents the rate of change of a function. Integral calculus, on the other hand, involves the accumulation of quantities and the areas under curves.
In our exercise, we applied differential calculus concepts by evaluating the limit of a function involving an exponential term. This involved expanding the function into its Maclaurin series and simplifying it to observe its behavior as \(x\) approached 0.
Calculus enables us to handle complex mathematical problems by breaking them down into simpler, more manageable parts, as seen in the step-by-step solution provided.

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Most popular questions from this chapter

Find the Maclaurin series for the following functions. $$e^{1-\sqrt{1-x^{2}}}$$

A tall tower of circular cross section is reinforced by horizontal circular disks (like large coins), one meter apart and of negligible thickness. The radius of the disk at height \(n\) is \(1 /(n \ln n)(n \geq 2)\) Assuming that the tower is of infinite height: (a) Will the total area of the disks be finite or not? Hint: Can you compare the series with a simpler one? (b) If the disks are strengthened by wires going around their circumferences like tires, will the total length of wire required be finite or not? (c) Explain why there is not a contradiction between your answers in (a) and (b). That is, how is it possible to start with a set of disks of finite area, remove a little strip around the circumference of each, and get an infinite total length of these strips? Hint: Think about units-you can't compare area and length. Consider two cases: (1) Make the width of each strip equal to one percent of the radius of the disk from which you cut it. Now the total length is infinite but what about the total area? (2) Try to make the strips all the same width; what happens? Also see Chapter 5, Problem 3.31(b).

Find the following limits using Maclaurin series and check your results by computer. Hint: First combine the fractions. Then find the first term of the denominator series and the first term of the numerator series. $$\text { (a) } \lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right)$$ $$\text { (b) } \lim _{x \rightarrow 0}\left(\frac{1}{x^{2}}-\frac{\cos x}{\sin ^{2} x}\right)$$ $$\text { (c) } \lim _{x \rightarrow 0}\left(\csc ^{2} x-\frac{1}{x^{2}}\right)$$ $$\text { (d) } \lim _{x \rightarrow 0}\left(\frac{\ln (1+x)}{x^{2}}-\frac{1}{x}\right)$$

Use the comparison test to prove the convergence of the following series: (a) \(\sum_{n=1}^{\infty} \frac{1}{2^{n}+3^{n}}\) (b) \(\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}\)

Use series you know to show that: $$\ln 3+\frac{(\ln 3)^{2}}{2 !}+\frac{(\ln 3)^{3}}{3 !}+\cdots=2$$

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