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Chapter 2: Problem 15

Use the results of this section to evaluate the limit. $$ \lim _{x \rightarrow 0} \ln \left(\frac{e^{x}-1}{x}\right) $$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Identify the form

Notice that as \( x \to 0 \), both the numerator \( e^x - 1 \) and the denominator \( x \) approach 0. Thus the limit is initially in the indeterminate form \( \frac{0}{0} \).
02

Expand the expression using series

Use the series expansion for \( e^x \), which is \( e^x = 1 + x + \frac{x^2}{2} + \cdots \). Substituting, we get:\[\frac{e^x - 1}{x} = \frac{1 + x + \frac{x^2}{2} - 1}{x} = \frac{x + \frac{x^2}{2}}{x}.\]
03

Simplify the expression

Simplify the expression obtained from the expansion:\[\frac{x + \frac{x^2}{2}}{x} = 1 + \frac{x}{2}.\] Therefore, as \( x \to 0 \), \( \frac{e^x - 1}{x} \to 1.\)
04

Evaluate the natural log

The given limit requires finding \( \ln \left( \frac{e^x - 1}{x} \right) \). Substituting the limit found in step 3, as \( x \to 0 \):\[\ln \left( \frac{e^x - 1}{x} \right) \to \ln(1) = 0.\]
05

Conclusion

Conclude that the limit of the logarithm evaluates to 0 as \( x \) approaches 0. Therefore, \(\lim_{x \to 0} \ln \left( \frac{e^x - 1}{x} \right) = 0.\)

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Key Concepts

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

L'Hopital's Rule
When dealing with limits that result in indeterminate forms like \( \frac{0}{0} \), L'Hopital's Rule comes to the rescue. This rule is a tool in calculus that helps evaluate such limits by differentiating the numerator and the denominator. If after trying the substitution the limit still results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hopital's Rule can be applied again. Here鈥檚 how it works:
  • First, ensure the limit results in an indeterminate form like \( \frac{0}{0} \).
  • Differentiate the numerator and the denominator separately.
  • Find the limit of the new fraction.
This powerful rule simplifies complex limit problems where direct substitution fails, especially when functions have complex expressions.
Indeterminate Forms
Indeterminate forms occur in mathematics when substituting variables into a function results in an undefined form. In calculus, common indeterminate forms include \( \frac{0}{0} \), \( \infty - \infty \), \( 0 \times \infty \), and more. These forms are called indeterminate because they do not immediately provide enough information to determine the limit. Here鈥檚 why they matter:
  • Indeterminate forms indicate uncertainty in the behavior of a function at a point.
  • They require special techniques like L'Hopital's Rule or algebraic manipulation to resolve.
  • Understanding these forms allows us to approach complex limits methodically.
Recognizing and resolving indeterminate forms is crucial for accurate limit evaluation and understanding the underlying function behavior.
Series Expansion
Series expansion is a mathematical tool used to represent functions as the sum of an infinite series of terms. One familiar form is the Taylor series, which provides a polynomial approximation of functions around a certain point. Here's how series expansions are used in calculus:
  • The exponential function \( e^x \) can be expressed as \( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \).
  • This expansion is particularly useful for approximations when \( x \) is near 0, simplifying limits and derivatives.
  • Complex problems can be tackled by substituting the series into the function to simplify or investigate behavior.
Series expansion facilitates solving limits by breaking down functions into simpler components, making complex calculus problems more manageable.

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Most popular questions from this chapter

Determine whether \(f\) is continuous at \(a\). $$ f(x)=\frac{x^{4}+x^{2}-2}{x^{2}-1} ; a=0 $$

Explain why \(f\) is continuous on the given interval or intervals $$ f(x)=\left\\{\begin{array}{l} (\sin x) / x \text { for } x \neq 0 \\ 1 \text { for } x=0 \end{array} ;(-\infty, \infty)\right. $$

Explain why \(f\) is continuous on the given interval or intervals $$ f(x)=\tan \left(\frac{1}{x^{2}+1}\right) ;(-\infty, \infty) $$

Determine whether \(f\) is continuous or discontinuous at \(a\). If \(f\) is discontinuous at \(a\), determine whether \(f\) is continuous from the right at \(a\), continuous from the left at \(a\), or neither. $$ f(x)=\sqrt{x-2} ; a=2 $$

Show that the equation has at least one solution. $$ \begin{gathered} 2 x \\ x^{7 / 3}+x^{5 / 3}-1=0 \end{gathered} $$
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