Problem 17 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

\(\lim _{x \rightarrow 0} \frac{\sin ^{-1} x-\tan ^{-1} x}{x^{3}}\) is equal to (a) 0 (b) 1 (c) \(-1\) (d) \(1 / 2\)

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: 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.

Step 2: 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 \( x - \frac{x^3}{3} + O(x^5) \).

Step 3: Substitute Taylor Expansions into the Expression

Substitute the expansions from Step 2 into the expression \( \sin^{-1} x - \tan^{-1} x \). This gives us: \( (x + \frac{x^3}{6}) - (x - \frac{x^3}{3}) = \frac{x^3}{6} + \frac{x^3}{3} = \frac{x^3}{6} + \frac{2x^3}{6} = \frac{3x^3}{6} = \frac{x^3}{2} \).

Step 4: 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 \( \lim_{x \rightarrow 0} \frac{1}{2} = \frac{1}{2} \).

Step 5: 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 \( \frac{1}{2} \).

Key Concepts

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:

• 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 \( x \rightarrow 0 \).

By using Taylor series expansions, we can simplify inverse trigonometric expressions. This simplification makes it much easier to evaluate the limit. Think of it as making very complex functions more user-friendly!

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:

• \( \sin^{-1} x \) 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.

In our exercise, \( \sin^{-1} x \) and \( \tan^{-1} x \) are approximated at \( x \approx 0 \) using Taylor expansions. This approximation is essential to simplify the limit expression. It's like peeling away layers to reveal a simpler core.

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.

In the given problem, direct substitution leads nowhere productive. Using Taylor series to expand \( \sin^{-1} x \) and \( \tan^{-1} x \) simplifies the expression. This way, we can clearly see the terms \( x^3 \) cancel out. The limit simplifies into a direct and simple constant, \( \frac{1}{2} \). Think of it as ingeniously smoothening a bumpy road.

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:

• 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.

In cases involving limits as \( x \rightarrow 0 \), these approximations help avoid indefinite forms. They transform intricate functions into algebraic sums or polynomials. This transition is crucial in solving limits efficiently, especially for complex expressions like combinations of inverse trigonometric functions. You can think of asymptotic approximations as zooming out to see the bigger picture—enabling us to make sense of seemingly complicated and cumbersome expressions.