Evaluating Limits Involving Trigonometric and Polynomial Functions

In calculus, evaluating limits is a fundamental concept. Limits can be particularly challenging when they involve both trigonometric and polynomial functions. This article will guide you through the process using a specific example: lim_{xto 0}frac{sqrt{1x^3}-1}{1-cos xcdot arctan 5x}.

Introduction to Limits

There are several methods to evaluate limits, such as direct substitution, factorization, and the use of Taylor expansions. In this article, we will cover the Taylor expansion method, which is particularly useful for more complex functions.

Evaluating the Limit Using Taylor Expansions

The given limit is as follows:

Start{align} sqrt{1x^3}1 frac{x^3}{2}ox, cos x1-frac{x^2}{2}ox^2, arctan xx oxend{align}

To understand the steps, let’s break it down:

For the numerator: sqrt{1x^3}-11 frac{x^3}{2}ox^3-1frac{x^3}{2}ox^3. For the denominator: 1-cos xcdot arctan 5x(1-1 frac{x^2}{2}yx^2)cdot 5xx^2cdot 5x5x^3.

Substituting these into the original limit, we get:

Begin{align*} lim_{xto 0} frac{left1 frac{x^3}{2}ox^3right-1}{left1-1 frac{x^2}{2}ox^2right5xox}lim_{xto 0}frac{x^3/2ox^3}{5x^3/2ox^3} frac{1}{5}. End{align*}

Using the Conjugate to Rationalize the Numerator

Another approach to simplifying the limit is by rationalizing the numerator. This can be done by multiplying the numerator and the denominator by the conjugate of the numerator. The conjugate of sqrt{1x^3}-1 is sqrt{1x^3} 1. Let’s apply this:

Start{align} lim_{x to 0} frac{sqrt{1x^3}-1}{1-cos xcdot arctan 5x}cdot frac{sqrt{1x^3} 1}{sqrt{1x^3} 1} lim_{x to 0} frac{x^3}{1-cos xcdot arctan 5x(sqrt{1x^3} 1)} lim_{x to 0} frac{x^3}{1-cos xcdot 5x(sqrt{1x^3} 1)} lim_{x to 0} frac{x^3}{1-cos xcdot 5x(sqrt{1x^3} 1)} lim_{x to 0} frac{1}{sqrt{1x^3} 1}cdot frac{x}{5x}cdot frac{x^2}{1-cos x} 1/sqrt{10^3} 1cdot lim_{x to 0} frac{1}{5}cdot lim_{x to 0} frac{x^2}{1-cos x}. End{align}

By applying L'H?pital's Rule and using the fact that lim_{x to 0}frac{sin x}{x}1, we get:

1/sqrt{10^3} 1cdot 1/5cdot 2/11/5.

Final Answer

Using the conjugate to rationalize the numerator, we break the limit into three factors and evaluate each one:

Start{align} lim_{x to 0} frac{1}{sqrt{1x^3} 1}cdot lim_{x to 0} frac{x}{arctan 5x}cdot lim_{x to 0} frac{x^2}{1-cos x} 1/sqrt{10^3} 1cdot lim_{x to 0} 1arctan 5xcdot lim_{x to 0} frac{x^2}{1-cos x} 1/sqrt{10^3} 1cdot 1/5cdot lim_{x to 0} frac{2x}{sin x} 1/sqrt{10^3} 1cdot 1/5cdot 2/1 1/5. End{align}

This shows that the limit evaluates to 1/5.

Key Points

Understanding the given function and breaking it down into simpler parts is crucial. Taylor expansions provide a powerful tool for evaluating complex limits. The conjugate method helps in simplifying rational expressions. Applying L'H?pital's Rule appropriately can simplify the evaluation process.

In conclusion, evaluating the given limit using both Taylor expansions and the conjugate method both lead to the same result: 1/5. This demonstrates the versatility of these techniques in solving complex calculus problems.