/*! 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 2 Find the following limits if the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 2

Find the following limits if they exist. (a) \(\lim _{x \rightarrow 0} \frac{x^{3}}{\sin x-x}\) (b) \(\lim _{x \rightarrow 0} \frac{\tan x-x}{x^{3}}\) (c) \(\lim _{x \rightarrow 0}\left[\frac{1}{\sin x}-\frac{1}{x}\right]\) (d) \(\lim _{x \rightarrow 0}(\cos x)^{1 / x^{2}}\)

Short Answer

Expert verified
a) 0, b) 1/3, c) 1/2, d) 1/\sqrt{e}

Step by step solution

01

Evaluate limit (a)

For \(\lim _{x \rightarrow 0} \frac{x^{3}}{\sin x-x}\), apply L'H么pital's rule because both the numerator and the denominator approach 0. Take derivatives of the numerator and the denominator: \(\lim _{x \to 0} \frac{3x^{2}}{\cos x-1}\). Now repeat L'H么pital's rule again as it still forms 0/0: \(\lim _{x \to 0} \frac{6x}{-\sin x}\). This is equal to 0 because the numerator is 0.
02

Evaluate limit (b)

For \(\lim _{x \rightarrow 0} \frac{\tan x-x}{x^{3}}\), use the series expansion for \(\tan x\): \(\tan x \approx x + \frac{x^3}{3} + O(x^5)\). Thus, \(\tan x - x \approx \frac{x^3}{3}\). The limit becomes \(\lim _{x \to 0} \frac{\frac{x^3}{3}}{x^3} = \frac{1}{3}\).
03

Evaluate limit (c)

For \(\lim _{x \rightarrow 0}\left[\frac{1}{\sin x}-\frac{1}{x}\right]\), rewrite the expression as a common denominator: \(\lim _{x \to 0} \frac{x-\sin x}{x \sin x}\). Apply L'H么pital's rule by differentiating the numerator and the denominator: \(\lim _{x \to 0} \frac{1-\cos x}{\sin x + x \cos x}\). As \(x\) approaches 0, \(1 - \cos x \approx \frac{x^2}{2}\) and \(\sin x \approx x\). Simplifying, \(\lim _{x \to 0} \frac{\frac{x^2}{2}}{x^2} = \frac{1}{2}\).
04

Evaluate limit (d)

For \(\lim _{x \rightarrow 0}(\cos x)^{1 / x^{2}}\), take the natural logarithm and simplify: \(y = (\cos x)^{1/x^2}, \) then \(\ln y = \frac{\ln (\cos x)}{x^2}\). As \(x\) approaches 0, \(\cos x \approx 1 - \frac{x^2}{2}\), thus \(\ln (\cos x) \approx -\frac{x^2}{2}\). This makes the expression \(\lim _{x \to 0} \frac{-\frac{x^2}{2}}{x^2} = -\frac{1}{2}\), hence \(\ln y = -\frac{1}{2}\). Thus, \(y = e^{-1/2} = \frac{1}{\sqrt{e}}\).

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 91影视!

Key Concepts

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

L'H么pital's rule
L'H么pital's rule is a handy tool in calculus when you need to find the limit of a fraction that results in an indeterminate form, like 0/0 or 鈭�/鈭�.
This rule states that if \(\frac{f(x)}{g(x)}\) approaches an indeterminate form as x approaches a certain value, you can differentiate the numerator, f(x), and the denominator, g(x), and then find the limit of \(\frac{f'(x)}{g'(x)}\).
Repeat until you get a determinate form. Here's a simplified version of the rule:
If \(\frac{f(x)}{g(x)} = \frac{0}{0}\), then \(\frac{f'(x)}{g'(x)}\) may help find the limit.
If after taking derivatives, \(\frac{f'(x)}{g'(x)} = \frac{0}{0}\) again, continue applying the rule.
For example, consider \(\frac{x^3}{\text{sin}x - x}\). Applying L'H么pital's rule once, you get \(\frac{3x^2}{\text{cos}x - 1}\).
Since you still have an indeterminate form (0/0), apply the rule again to get \(\frac{6x}{-\text{sin}x}\).
Now, as x approaches 0, the limit becomes 0 because the numerator is 0.
This step-by-step application shows how L'H么pital's rule can simplify finding limits in calculus.
Series expansion
A series expansion is used to approximate functions by breaking them down into simpler terms.
It's particularly useful for functions like \(\text{tan}x\) or \(\text{sin}x\) near certain points.
Mathematically, these expansions are called Taylor or Maclaurin series.
For instance, the series expansion for \(\text{tan}x\) near 0 is:\( \text{tan}x = x + \frac{x^3}{3} + O(x^5)\).
Using this series expansion can help solve limits when direct substitution does not work.
For example, to evaluate \(\frac{\text{tan}x - x}{x^3}\) as x approaches 0, use the expansion for \(\text{tan}x\):
\(\text{tan}x \approx x + \frac{x^3}{3}\).
Hence, \(\text{tan}x - x \approx \frac{x^3}{3}\).
Substituting this back into the limit, you get \(\frac{\frac{x^3}{3}}{x^3} = \frac{1}{3}\).
Here, you see how series expansion simplifies the problem and provides a straightforward solution.
Natural logarithm
The natural logarithm, denoted as \( \text{ln}(x) \), is the logarithm to the base e, where e is an irrational constant approximately equal to 2.71828.
It plays a crucial role in calculus, particularly in limit problems.
For example, evaluating limits involving expressions like \( (\text{cos}x)^{\frac{1}{x^2}} \) can be tricky.
Here's how to break it down:
Take the natural logarithm of the expression and simplify.
If \( y = (\text{cos}x)^{\frac{1}{x^2}} \), then \( \text{ln}(y) = \frac{\text{ln}(\text{cos}x)}{x^2} \).
Near zero, \(\text{cos}x \approx 1 - \frac{x^2}{2}\), so \(\text{ln}(\text{cos}x) \approx -\frac{x^2}{2}\).
Substituting this approximation, we get \( \text{ln}(y) \approx \frac{-\frac{x^2}{2}}{x^2} = -\frac{1}{2} \).
Thus, \( y = e^{-\frac{1}{2}} = \frac{1}{\text{sqrt(e)}}\).
Using the natural logarithm simplifies the expression and makes evaluating the limit possible.
Understanding \( \text{ln}(x) \) and its properties is essential for tackling complex limit problems.

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

(a) Show that \(x<\tan x\) for all \(x \in\left(0, \frac{\pi}{2}\right)\) (b) Show that \(\frac{x}{\sin x}\) is a strictly increasing function on \(\left(0, \frac{\pi}{2}\right)\) (c) Show that \(x \leq \frac{\pi}{2} \sin x\) for \(x \in\left[0, \frac{\pi}{2}\right]\)

Let \(f\) be a function defined on an open interval \(I\) containing \(a\). Let \(h\) be a function defined on an open interval \(J\) containing \(f(a)\) except at \(f(a),\) and suppose that \(f(x) \in J\) and \(f(x) \neq f(a)\) for all \(x \in I \backslash\\{a\\} .\) Then \(h \circ f\) is defined on \(I \backslash\\{a\\} .\) Use Corollary 20.7 to prove that if \(\lim _{x \rightarrow a} f(x)=f(a)\) and if \(\lim _{y \rightarrow f(a)} h(y)\) exists and is finite, then \(\lim _{x \rightarrow a} h \circ f(x)=\lim _{y \rightarrow f(a)} h(y)\)

Find the Taylor series for \(\cos x\) and indicate why it converges to \(\cos x\) for all \(x \in \mathbb{R}\)

Suppose that \(f\) is differentiable on \(\mathbb{R},\) that \(1 \leq f^{\prime}(x) \leq 2\) for \(x \in \mathbb{R}\) and that \(f(0)=0 .\) Prove that \(x \leq f(x) \leq 2 x\) for all \(x \geq 0\)

Let \(f\) be differentiable on \(\mathbb{R}\) with \(a=\sup \left\\{\left|f^{\prime}(x)\right|: x \in \mathbb{R}\right\\}<1\) Select \(s_{0} \in \mathbb{R}\) and define \(s_{n}=f\left(s_{n-1}\right)\) for \(n \geq 1 .\) Thus \(s_{1}=f\left(s_{0}\right)\) \(s_{2}=f\left(s_{1}\right),\) etc. Prove that \(\left(s_{n}\right)\) is a convergence sequence. Hint: To show \(\left(s_{n}\right)\) is Cauchy, first show that \(\left|s_{n+1}-s_{n}\right| \leq a\left|s_{n}-s_{n-1}\right|\) for \(n \geq 1\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation
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.