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Chapter 7: Problem 28

Use l'Hopital's rule to find the limits in Exercises \(7-50\) $$\lim _{\theta \rightarrow 0} \frac{(1 / 2)^{\theta}-1}{\theta}$$

Short Answer

Expert verified
The limit is \(-\ln(2)\).

Step by step solution

01

Identify the Indeterminate Form

First, we need to evaluate the limit directly to check if it forms an indeterminate form. As \( \theta \to 0 \), both the numerator \( (1/2)^\theta - 1 \) and the denominator \( \theta \) approach 0. Therefore, we have the indeterminate form \( \frac{0}{0} \), which allows us to use l'Hopital's Rule.
02

Apply l'Hopital's Rule

l'Hopital's Rule states that for limits of indeterminate forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can find the limit by differentiating the numerator and the denominator separately:\[\lim _{\theta \rightarrow 0} \frac{(1 / 2)^{\theta}-1}{\theta} = \lim _{\theta \rightarrow 0} \frac{\frac{d}{d\theta}((1/2)^\theta - 1)}{\frac{d}{d\theta}(\theta)}.\]
03

Differentiate the Numerator and Denominator

Differentiate the numerator, \( (1/2)^\theta \), using the chain rule. The derivative is:\[-\ln(2) \cdot (1/2)^\theta.\]The derivative of the denominator \( \theta \) is simply 1.
04

Reevaluate the Limit Using Derivatives

Substitute the derivatives into the limit expression:\[\lim _{\theta \rightarrow 0} \frac{-\ln(2) \cdot (1/2)^\theta}{1} = -\ln(2) \cdot (1/2)^0 = -\ln(2) \cdot 1 = -\ln(2).\]
05

Conclusion of the Calculation

The limit of the given expression as \( \theta \to 0 \) is calculated as \(-\ln(2)\) after applying l'Hopital's Rule and simplifying the expression.

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Key Concepts

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

Indeterminate Forms
When calculating limits, you might encounter something called an indeterminate form. This happens when both the numerator and the denominator of a fraction approach specific values that lead to an undefined mathematical expression. The most common types of indeterminate forms are \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). These forms are crucial in calculus because they signal that normal limit calculation methods won't work directly.

  • For example, if you have \( \lim_{x \to a} \frac{f(x)}{g(x)} \) where \( f(a) = 0 \) and \( g(a) = 0 \), you end up with \( \frac{0}{0} \).
  • Another indeterminate form is \( \frac{\infty}{\infty} \) when both the numerator and denominator tend toward infinity as \( x \) approaches a certain value.
When you identify an indeterminate form, l'Hopital's Rule often comes to the rescue by providing a method to find the limit despite the form's complexity.
Differentiation
Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which tells us the rate at which the function's value changes. Understanding differentiation is key when using l'Hopital's Rule, as the rule requires taking derivatives of both the numerator and the denominator of a fraction.

The basic idea of differentiation involves:
  • The derivative of a function \( f(x) \) is the function \( f'(x) \).
  • It measures how \( f(x) \) changes as \( x \) changes.
  • It is often represented as \( \frac{dy}{dx} \), where \( y = f(x) \).
When applying l'Hopital's Rule, you differentiate the top and the bottom separately. When you've got more complicated expressions, like \((1/2)^\theta\), using the chain rule becomes necessary. The chain rule helps differentiate complex functions by breaking them down into simpler parts.
Limits in Calculus
Limits are foundational to understanding calculus. They describe the behavior of a function as it approaches a certain point. The limit helps to explore how functions behave near points where they aren't explicitly defined, and is crucial for finding the exact trends of functions as one variable nears a particular value.

In calculus, understanding limits means grasping:
  • The symbol \( \lim_{x \to a} f(x) \) represents the value that \( f(x) \) approaches as \( x \) gets closer to \( a \).
  • It allows us to define derivatives and integrals, two of the main operations in calculus.
  • Helps solve problems where direct substitution doesn't work due to indeterminate forms or undefined values.
Using l'Hopital's Rule involves evaluating limits, especially when straightforward methods don't yield results due to indeterminate forms. This rule provides a powerful way to discover what value a function approaches by shifting the problem from algebraic limits to derivatives.

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Most popular questions from this chapter

Evaluate the integrals in Exercises \(67-74\) in terms of a. inverse hyperbolic functions. b. natural logarithms. $$\int_{1}^{e} \frac{d x}{x \sqrt{1+(\ln x)^{2}}}$$

The continuous extension of \((\sin x)^{x}\) to \([0, \pi]\) \begin{equation} \begin{array}{l}{\text { a. Graph } f(x)=(\sin x)^{x} \text { on the interval } 0 \leq x \leq \pi . \text { What }} \\\\\quad {\text { value would you assign to } f \text { to make it continuous at } x=0 ?} \\ {\text { b. Verify your conclusion in part (a) by finding } \lim _{x \rightarrow 0^{\prime}} f(x)} \\\\\quad {\text { with I'Hopital's Rule. }}\\\\{\text { c. Returning to the graph, estimate the maximum value of } f \text { on }} \\ \quad{[0, \pi] \text { . About where is max } f \text { taken on? }}\\\\{\text { d. Sharpen your estimate in part (c) by graphing } f^{\prime} \text { in the same }} \\\ \quad{\text { window to see where its graph crosses the } x \text { -axis. To simplify }} \\\\\quad {\text { your work, you might want to delete the exponential factor from }} \\\\\quad {\text { the expression for } f^{\prime} \text { and graph just the factor that has a zero. }}\end{array} \end{equation}

Find the limits in Exercises \(91-98\) $$ \lim _{x \rightarrow \infty} x \tan ^{-1} \frac{2}{x} $$

Find the limits in Exercises \(91-98\) $$\lim _{x \rightarrow 0} \frac{\sin ^{-1} 5 x}{x}$$

Use the formulas in the box here to express the numbers in Exercises \(61-66\) in terms of natural logarithms. $$\operatorname{csch}^{-1}(-1 / \sqrt{3})$$
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