Chapter 2: Problem 17
\(\lim _{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{x^{3}}\) is equal to (a) 0 (b) 1 (c) \(-1\) (d) \(1 / 2\)
Short Answer
Expert verified
The limit is \( \frac{1}{2} \).
Step by step solution
01
Identify the Limit Expression
The problem requires us to compute the limit \( \lim _{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{x^{3}} \). This involves simplifying the given expression to a form where the limit can be evaluated.
02
Use Taylor Expansions
We approximate \( \sin^{-1} x \) and \( \tan^{-1} x \) using their Taylor series expansions around \( x = 0 \). The leading term in \( \sin^{-1} x \) is \( x + \frac{x^3}{6} + O(x^5) \), and in \( \tan^{-1} x \), it is \( x - \frac{x^3}{3} + O(x^5) \).
03
Substitute Taylor Expansions into the Expression
Substitute the expansions from Step 2 into the expression \( \sin^{-1} x - \tan^{-1} x \). This gives us: \( (x + \frac{x^3}{6}) - (x - \frac{x^3}{3}) = \frac{x^3}{6} + \frac{x^3}{3} = \frac{x^3}{6} + \frac{2x^3}{6} = \frac{3x^3}{6} = \frac{x^3}{2} \).
04
Simplify the Limit Expression
Replace the numerator in the original limit expression with \( \frac{x^3}{2} \): \( \lim_{x \rightarrow 0} \frac{(\frac{x^3}{2})}{x^3} \). Simplifying this, we have \( \lim_{x \rightarrow 0} \frac{1}{2} = \frac{1}{2} \).
05
Evaluate the Limit
The limit value is constant because all \(x^3\) terms cancel out, leaving us with just \( \frac{1}{2} \). Thus, the limit is \( \frac{1}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series Expansion
The Taylor series expansion is a super powerful mathematical tool. It allows us to express complex functions as infinite sums of polynomial terms. This is especially useful near a specific point, often zero. Do you sometimes find functions too complex to work with directly? Taylor series offer a simpler way by approximating these functions.
Here's how it works:
Here's how it works:
- The series is centered around a certain point, usually zero.
- For inverse trigonometric functions like \( \sin^{-1} x \) and \( \tan^{-1} x \), these series help in breaking down their complexity.
- The leading terms are crucial when evaluating limits, especially when \( x \rightarrow 0 \).
Inverse Trigonometric Functions
Inverse trigonometric functions are the opposite of trigonometric functions. Their primary goal is to find the angle when given a trigonometric ratio. While ordinary trigonometric functions deal with circles and triangles, these inverse functions serve an inverse purpose.
Key points to understand:
Key points to understand:
- \( \sin^{-1} x \) often called arcsine, gives the angle whose sine is \( x \).
- \( \tan^{-1} x \), or arctangent, finds the angle whose tangent is \( x \).
- These functions are critical in solving problems where angles need finding.
- In calculus, they often appear in problems involving limits and derivatives.
Limit Evaluation Techniques
Limits in calculus signify the behavior of a function as it approaches a specific value or point. Evaluating limits effectively can be tricky, especially when direct substitution results in undefined forms, like \( \frac{0}{0} \). However, several techniques are at our disposal:
- Substitution: The simplest approach, useful when expressions aren't problematic.
- Taylor Expansion: Especially when functions involve complex forms like inverse trigonometric functions.
- Factoring and Rationalization: Helps in canceling problematic terms.
Asymptotic Approximations
Asymptotic approximations deal with the behavior of functions as they approach certain limits. They provide a way to glimpse the eventual always-toward nature of functions—particularly when these functions head towards infinity, zero, or any singular point.
Consider this:
Consider this:
- For small values of \( x \), asymptotic approximations offer simpler expressions.
- These approximations help keep calculations manageable and comprehensible.
- They can be based on Taylor series or other techniques, focusing on dominant terms.