Use Maclaurin series to evaluate: \(\lim _{x \rightarrow 0}\left(\frac{1}{x^{2}}-\cot ^{2} x\right)\)

Short Answer

Expert verified
\[\lim_{x \to 0} \left( \frac{1}{x^2} - \text{cot}^2 x \right) = \frac{2}{3} \]

Step by step solution

01

Recall the Maclaurin series for relevant functions

The Maclaurin series for \(\tan x\) is \[ x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + \dots \]and the Maclaurin series for \( \tan^{-1} x\) (also written as \( \text{arctan} x \)) is\[ x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \dots \]
02

Rewrite \( \text{cot}^2 x \)

Using the relation \( \text{cot} x = \frac{1}{\tan x} \), we need the series for \( \tan x \) and then take the reciprocal for \( \text{cot} x \). Given the Maclaurin series for small values of \( x \), we can write \( \tan x \) as approximately \( x + \frac{x^{3}}{3} \). Thus, \( \text{cot} x \) for small \( x \) is approximately \( \frac{1}{x + \frac{x^{3}}{3}} \).
03

Simplify \( \text{cot} x \) using series

For small \( x \, \text{cot} x \) can be approximated by simplifying the reciprocal function: \[ \text{cot} x ≈ \frac{1}{x} \left(1 - \frac{x^{2}}{3}\right) \]Therefore, \(\text{cot}^2 x\) becomes \[ \left( \frac{1}{x} (1 - \frac{x^{2}}{3}) \right)^2 ≈ \frac{1}{x^2} \left( 1 - \frac{2x^{2}}{3} + O(x^4) \right) \]
04

Subtract to find the limit expression

The limit expression is \( \frac{1}{x^2} - \text{cot}^2 x \). Substituting the approximation from Step 3: \[ \frac{1}{x^2} - \frac{1}{x^2} \left( 1 - \frac{2x^2}{3} + O(x^4) \right) \= \frac{1}{x^2} - \frac{1}{x^2} + \frac{2}{3} + O(x^2) \]As \( x \) approaches 0, the higher-order terms (\( O(x^2) \)) vanish, leaving \[ \lim_{x \to 0} \left( \frac{1}{x^2} - \text{cot}^2 x \right) = \frac{2}{3} \]

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 Vaia!

Key Concepts

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

Limit Evaluation
In mathematical analysis, evaluating limits is a fundamental concept that helps us understand the behavior of functions as they approach a specific point. For the given problem, we want to find \(\text{lim}_{x \to 0}\bigg(\frac{1}{x^2} - \cot^2 x \bigg)\). By using the Maclaurin series and simplifying the expressions, we can break down the process effectively:
  • First, we recall the Maclaurin series expansion for relevant functions such as the tangent, i.e., \(\tan x \) and its reciprocal form \(\cot x \).
  • By expressing \(\text{cot } x\) in its series form, this enables us to approximate it using simpler terms, significantly simplifying our task of evaluating the limit.
  • Finally, we subtract and simplify to find the specific value as \( x\) approaches 0.
This helps turn a complex limit problem into something very manageable.
Series Approximation
Series approximations are powerful tools in calculus for simplifying functions. The Maclaurin series is a specific type of power series that expands functions around 0. For example, the Maclaurin series for \( \tan x \) is given by:
\[ x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + \dots \]
Using this series, we approximate \( \tan x \) for values of \( x \) close to zero. Likewise, the inverse tangent or arctan has a series:
\[ x - \frac{x^3}{3} + \frac{x^5}{5} - \dots\]
These series are particularly useful when evaluating limits because they break down complex trigonometric functions into polynomials, which are easier to handle. In the given exercise, after obtaining the series for \( \tan x \), we evaluate \( \cot x \) by taking the reciprocal, and further, \( \cot^2 x \) by squaring the simplified expression. This process shows the power and utility of series approximations in providing a quick pathway to solving limit problems.
Trigonometric Functions
Trigonometric functions such as \( \sin, \cos,\tan,\text{ and } \cot\), are central to many areas of mathematics. They describe relationships in triangles and are used extensively in calculus. For instance:
  • The function \( \tan x \) relates the sine and cosine functions by \( \tan x = \frac{\sin x}{\cos x} \).
  • The function \( \cot x \), in turn, is the reciprocal of \( \tan x \).
These relationships allow mathematicians to use one trigonometric function to find others.
When given the task to evaluate limits involving trigonometric functions, approximations become very useful, particularly with functions like \( \tan x \) and \( \cot x \) which do not provide simple values when substituted directly. The Maclaurin series allows for a polynomial approximation of these functions, enabling simpler arithmetic operations. This process is elegantly demonstrated in our given exercise, where finding \( \lim_{x \to 0} (\frac{1}{x^2} - \cot^2 x) \) becomes straightforward after expressing \( \cot x \) in terms of its series approximation.

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

Test the following series for convergence using the comparison test. (a) \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) Hint: Which is largcr, \(n\) or \(\sqrt{n}\) ? (b) \(\sum_{n=2}^{\infty} \frac{1}{\ln n}\)

Use Maclaurin series to evaluate: \(\quad e^{-x, 6}-\left.\frac{1}{\sqrt{x}} \sin \sqrt{x}\right|_{x=0.0001}\)

Test the following series for convergence. \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{\ln n}\)

\(\sum_{n=2}^{\infty} \frac{1}{n^{2}-n}\)

Show that it is possible to stack a pile of identical books so that the top book is as far as you tike to the right of the bottom book. You may use as many books as you need, but for any desired distance it will be some finite number. Hint: Start at the top of the pile to see how to stack them. Each time place the pile already completed on top of another book so that the pile is just at the point of tipping. (In practice, of course, you can't let them overhang quite this much without having the stack topple.) Find the distance from the right-hand end of each book to the right-hand end of the one beneath it. (To find a general formula for this distance, consider the three forces acting on book \(n\), and write the equation for the torque about its right-hand end.) Show that the sum of these distances is a divergent series. (It is said that some students studying in the library tried this experiment one evening and lefe the results, to the consternation of the librarian the next morning. It is suggested that you try it instead with one or more decks of cards. Also see "Leaning Tower of The Physical Reciens," American Journal of Physics, vol. 27, no. 2, p. 12 t.)

See all solutions

Recommended explanations on Combined Science 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.

Sign-up for free