Chapter 1: Problem 27
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 \[ \lim_{x \to 0} \left( \frac{1}{x^2} - \text{cot}^2 x \right) = \frac{2}{3} \]
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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.
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.
\[ x + \frac{x^{3}}{3} + \frac{2x^{5}}{15} + \dots \]
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:
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.
- 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 \( \tan x \).
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.