Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 0} \frac{x-\sin 2 x}{\tan x} $$

Step-by-Step Solution

Answer

The limit is -1.

Step 1: Analyze the Limit

We need to find \( \lim_{x \to 0} \left( \frac{x - \sin 2x}{\tan x} \right) \). Both the numerator and denominator approach zero as \( x \to 0 \), which is an indeterminate form of \( \frac{0}{0} \). In such cases, L'Hôpital's Rule can be applied but let's simplify the expression using Taylor expansions first.

Step 2: Taylor Expansion of Sine and Tangent

The Taylor series expansions around \( x = 0 \) for \( \sin 2x \) and \( \tan x \) are useful here. For \( \sin 2x \), the expansion is \( 2x - \frac{(2x)^3}{6} + \cdots = 2x - \frac{8x^3}{6} + \cdots \). For \( \tan x \), the expansion is \( x + \frac{x^3}{3} + \cdots \). We'll substitute these into the expression to simplify.

Step 3: Simplify the Expression using Taylor Expansions

Substitute the expansions into the limit expression:\[\frac{x - (2x - \frac{8x^3}{6})}{x + \frac{x^3}{3}}= \frac{x - 2x + \frac{8x^3}{6}}{x + \frac{x^3}{3}}= \frac{-x + \frac{4x^3}{3}}{x + \frac{x^3}{3}}\]This simplifies to:\[ \frac{-1 + \frac{4x^2}{3}}{1 + \frac{x^2}{3}} \]

Step 4: Evaluate the Limit

Now, consider the limit as \( x \to 0 \):\[ \lim_{x \to 0} \frac{-1 + \frac{4x^2}{3}}{1 + \frac{x^2}{3}} = \frac{-1 + 0}{1 + 0} = -1 \]Thus, the limit simplifies directly to \(-1\).

Key Concepts

Indeterminate Forms

When solving limits in calculus, you may encounter expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are called indeterminate forms. They do not have a clear value at first glance. The expressions can lead to many different values depending on how their components behave as they approach certain points. This means we need extra analysis or rules to determine their limits.

• In the given problem, \( x \to 0 \) makes both the numerator and denominator equal to zero, giving an indeterminate form \( \frac{0}{0} \).
• Special techniques like L'Hôpital's Rule or expansions such as the Taylor series are often used to resolve these forms.

Taylor Series

The Taylor series offers a way to represent complex functions as infinite sums of polynomials. This makes it easier to perform calculations or simplify functions around a certain point, typically zero, known as the function's center. The series expansion provides approximations that are increasingly accurate as you add more terms.

• The formula for \( \sin x \) is \( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \).
• For \( \tan x \), it's \( x + \frac{x^3}{3} + \ldots \).

In the exercise, we use these approximations to transform the indeterminate form into something we can analyze more easily. By substituting Taylor series expansions for \( \sin 2x \) and \( \tan x \), the expression simplifies into fractions that can be evaluated as limits approach zero.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for evaluating limits with indeterminate forms like \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). This rule says if you have an indeterminate form, the limit of the ratio of the functions is the same as the limit of the ratio of their derivatives, provided that these limits exist.

• In the exercise above, L'Hôpital's Rule could be used since the expression initially forms \( \frac{0}{0} \).
• This would involve differentiating both the numerator and the denominator before retrying to evaluate the limit.

However, in our solution, we simplified using Taylor series instead, showing an alternative and often convenient approach to resolving these dilemmas.

Trigonometric Limits

Trigonometric limits often occur in calculus, especially when dealing with functions like sine, cosine, and tangent as they approach zero or infinity. These can sometimes form indeterminate expressions.

• One common limit is \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).
• Similarly, \( \tan x \) around zero can be tricky but is quite approachable with Taylor series expansion.

In such cases, understanding the quick behavior of these trigonometric functions around certain values is essential. Always look for patterns or memorized limits that might resolve the calculations more easily, as seen in the Taylor expansions and simplified expressions in the exercise.