/*! 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 20 Use l'Hopital's rule to find the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use l'Hopital's rule to find the limits in Exercises \(7-50\) $$\lim _{x \rightarrow 1} \frac{x-1}{\ln x-\sin \pi x}$$

Short Answer

Expert verified
The limit is \(\frac{1}{1+\pi}\).

Step by step solution

01

- Check the Indeterminate Form

First, we need to check if the expression is in the indeterminate form \(\frac{0}{0}\). Substitute \(x = 1\) into the numerator and the denominator: \(x - 1 = 1 - 1 = 0\), \(\ln 1 = 0\), and \(\sin \pi \cdot 1 = \sin \pi = 0\). Hence, \(\ln 1 - \sin \pi \cdot 1 = 0 - 0 = 0\). Thus, the expression \(\frac{x-1}{\ln x-\sin \pi x}\) is in the \(\frac{0}{0}\) indeterminate form.
02

- Apply l'Hopital's Rule

Since the limit is in the \(\frac{0}{0}\) indeterminate form, we can apply l'Hopital's Rule, which allows us to take the derivative of the numerator and the derivative of the denominator separately: \(\lim _{x \rightarrow 1} \frac{x-1}{\ln x - \sin \pi x} = \lim _{x \rightarrow 1} \frac{d/dx(x-1)}{d/dx(\ln x - \sin \pi x)}\).
03

- Differentiate the Numerator and Denominator

Differentiate the numerator: \(\frac{d}{dx}(x-1) = 1\).Differentiate the denominator: \(\frac{d}{dx}(\ln x - \sin \pi x) = \frac{1}{x} - \pi \cos \pi x\).
04

- Evaluate the New Limit

Substitute \(x = 1\) into the derivatives: \(\lim_{x \rightarrow 1} \frac{1}{\frac{1}{x} - \pi \cos \pi x} = \frac{1}{\frac{1}{1} - \pi \cos \pi \cdot 1}\).Calculate individually: \(1 - \pi \cos \pi = 1 - \pi (-1) = 1 + \pi\). So, the limit is \(\frac{1}{1 + \pi}\).

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.

Indeterminate Forms
Indeterminate forms occur when evaluating a limit leads to an ambiguous result, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These results don't provide clear information about the limit itself. To determine a concrete limit value, one must use additional techniques, such as factoring, simplifying, or applying calculus rules like L'Hôpital's Rule.

In the given example, substituting \(x = 1\) in both the numerator \(x-1\) and denominator \(\ln x - \sin \pi x\) gives \(0/0\), an classic indeterminate form. This is a cue to apply L'Hôpital's Rule.

L'Hôpital's Rule specifically addresses indeterminate forms like \(\frac{0}{0}\) by allowing differentiation of the numerator and denominator separately until a determinate result is found.
Calculus Limits
In calculus, limits help us understand the behavior of functions as inputs approach certain values. They form the foundation for defining continuity, derivatives, and integrals. To use limits effectively, one needs to understand how to manipulate expressions to bypass undefined or indeterminate forms.

Limits can describe several scenarios, such as:
  • An expression approaching a finite number: \(\lim_{x \to a} f(x) = L\).
  • An expression becoming infinitely large or small.
  • Behavior of functions as \(x\) approaches infinity.

In our exercise, the limit \(\lim_{x \to 1} \frac{x-1}{\ln x - \sin \pi x}\) aims to determine the behavior of this expression as \(x\) approaches 1. By reformulating it using L'Hôpital's Rule, we change the expression into something easier to evaluate.
Derivative Applications
Derivatives are key tools in calculus, helping us understand how functions change. They represent the rate of change of a function, providing insights into the function's behavior at specific points.

In the context of L'Hôpital's Rule, derivatives simplify the evaluation of limits by transforming indeterminate forms into more workable expressions.
  • The derivative of the numerator \(x-1\) is straightforward: it is 1.
  • For the denominator, \(\ln x - \sin \pi x\), the derivative involves applying derivative rules to each term: \(\frac{1}{x} - \pi \cos(\pi x)\).

Substituting back into the limit, these derivatives replace the original expression, giving \(\lim_{x \to 1} \frac{1}{\frac{1}{x} - \pi \cos \pi x}\), which resolves to \(\frac{1}{1+\pi}\), a clear, determinate value.

One App. One Place for Learning.

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

Get started for free

Most popular questions from this chapter

Evaluate the integrals in Exercises \(67-74\) in terms of a. inverse hyperbolic functions. b. natural logarithms. $$\int_{5 / 4}^{2} \frac{d x}{1-x^{2}}$$

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}

Evaluate the integrals in Exercises \(41-60\) $$\int_{-\ln 2}^{0} \cosh ^{2}\left(\frac{x}{2}\right) d x$$

Use the identity $$\csc ^{-1} u=\frac{\pi}{2}-\sec ^{-1} u$$ to derive the formula for the derivative of \(\csc ^{-1} u\) in Table 7.3 from the formula for the derivative of \(\sec ^{-1} u .\)

Evaluate the integrals in Exercises \(81-90\) $$ \int \frac{1}{\sqrt{x}(x+1)\left(\left(\tan ^{-1} \sqrt{x}\right)^{2}+9\right)} d x $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.