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Chapter 1: Problem 26

Apply the result if needed to find the limits. $$\lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x}$$

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Understand the Problem

We need to find the limit \( \lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x} \). As \( x \to 0 \), both the numerator \( \tan^{-1} x \) and the denominator \( x \) approach 0, creating an indeterminate form \( \frac{0}{0} \). We need to resolve this indeterminacy to find the limit.
02

Recall Taylor Series Expansion

Recall the Taylor series expansion for \( \tan^{-1} x \) around \( x = 0 \):\[ \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \].For small \( x \), the dominant term is \( x \), allowing us to approximate \( \tan^{-1} x \approx x \).
03

Simplify the Expression Using a Series

Using the series expansion, replace \( \tan^{-1} x \) with its first term \( x \), giving us:\[ \frac{\tan ^{-1} x}{x} \approx \frac{x}{x} = 1 \].As \( x \to 0 \), higher-order terms become negligible, making the approximation more accurate.
04

Evaluate the Limit

Thus, the limit evaluates to:\[ \lim _{x \rightarrow 0} \frac{\tan ^{-1} x}{x} = 1 \].

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Key Concepts

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

Indeterminate Forms
In calculus, an indeterminate form arises when evaluating a limit produces an undefined expression, often looking like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Such forms are initially ambiguous, because they do not directly give us the limit's value.

When we see \( \frac{0}{0} \), we must apply some algebraic or calculus techniques to resolve the indeterminacy and find the actual limit.
  • L'H么pital's Rule: For certain cases, if both the numerator and the denominator approach zero, this rule allows us to take derivatives of the numerator and denominator to find the limit.
  • Series Expansion: We can also use Taylor series to re-express functions near the point of interest, thus simplifying the evaluation of limits.
Understanding indeterminate forms is critical for solving limits, especially in expressions involving algebraic manipulation or transcendental functions like trigonometric and logarithmic functions.
Taylor Series
Taylor series is a powerful tool in calculus used to approximate functions. It expands a function into an infinite sum of terms calculated from the function's derivatives at a single point.

This becomes invaluable for evaluating limits when direct substitution results in an indeterminate form.
  • For example, the Taylor series expansion of \( \tan^{-1} x \) around \( x = 0 \) is given by: \[ \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \]
  • In our exercise, using the series allowed us to approximate \( \tan^{-1} x \approx x \) for small values of \( x \), where higher-order terms become negligible.
  • This approximation simplifies the limit calculation by essentially turning it into calculable terms, avoiding direct indeterminate form.
Taylor series is a go-to method for solving limits due to its ability to express tricky functions in a simpler polynomial form.
Trigonometric Functions
Trigonometric functions, like \( \tan^{-1} x \), play a prominent role in calculus, particularly when dealing with angles and periodic phenomena.

Inverse trigonometric functions, such as the arctan (or \( \tan^{-1} \)), help find angles given the trigonometric ratio.
  • As \( x \) approaches zero, \( \tan^{-1} x \) approaches zero as well, but the way it does so can lead to an indeterminate form.
  • This is why the Taylor series becomes necessary to break down \( \tan^{-1} x \) into more calculable terms when evaluating limits.
Understanding how trigonometric functions behave can greatly assist in making sense of limits involving these functions, especially under infinitesimally small or large transformations.

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Most popular questions from this chapter

Suppose that \(f\) is continuous on the interval [0,1] and that \(0 \leq f(x) \leq 1\) for all \(x\) in this interval. (a) Sketch the graph of \(y=x\) together with a possible graph for \(f\) over the interval [0,1]. (b) Use the Intermediate-Value Theorem to help prove that there is at least one number \(c\) in the interval [0,1] such that \(f(c)=c\).

According to one source, the noise inside a moving automobile is about \(70 \mathrm{dB},\) whereas an electric blender generates 93 dB. Find the ratio of the intensity of the noise of the blender to that of the automobile.

Let $$ f(x)=\left\\{\begin{array}{ll} \frac{x^{2}-9}{x+3}, & x \neq-3 \\ k, & x=-3 \end{array}\right. $$ (a) Find \(k\) so that \(f(-3)=\lim _{x \rightarrow-3} f(x)\) (b) With \(k\) assigned the value \(\lim _{x \rightarrow-3} f(x),\) show that \(f(x)\) can be expressed as a polynomial.

(a) Find the smallest positive number \(N\) such that for each \(x\) in the interval \((N,+\infty),\) the value of the function \(f(x)=1 / x^{2}\) is within 0.1 unit of \(L=0\) (b) Find the smallest positive number \(N\) such that for each \(x\) in the interval \((N,+\infty),\) the value of \(f(x)=\) \(x /(x+1)\) is within 0.01 unit of \(L=1\) (c) Find the largest negative number \(N\) such that for each \(x\) in the interval \((-\infty, N)\), the value of the function \(f(x)=1 / x^{3}\) is within 0.001 unit of \(L=0\) (d) Find the largest negative number \(N\) such that for each \(x\) in the interval \((-\infty, N),\) the value of the function \(f(x)=x /(x+1)\) is within 0.01 unit of \(L=1\)

If one sound is three times as intense as another, how much greater is its decibel level?
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