Problem 16 - Calculus for Biology and Medicine

Question

Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{1-\cos (2 x)}{3 x} $$

Step-by-Step Solution

Answer

The limit is \( \frac{2}{3} \).

Step 1: Identify the Limit to Evaluate

The given limit to evaluate is \( \lim _{x \rightarrow 0} \frac{1-\cos (2 x)}{3 x} \). Our goal is to find its value as \( x \) approaches 0.

Step 2: Use Trigonometric Identity

Recall the trigonometric identity \( 1 - \cos(\theta) = 2 \sin^2(\theta/2) \). Here, substitute \( \theta = 2x \), which leads to \( 1 - \cos(2x) = 2 \sin^2(x) \).

Step 3: Substitute the Identity in the Original Expression

Substitute \( 1 - \cos(2x) = 2\sin^2(x) \) into the expression: \( \lim _{x \rightarrow 0} \frac{2\sin^2(x)}{3x} \). The limit now becomes \( \lim _{x \rightarrow 0} \frac{2 \sin^2(x)}{3x} \).

Step 4: Rewrite to Simplify

Rewrite the limit as \( \lim _{x \rightarrow 0} \frac{2 \sin^2(x)}{x} \cdot \frac{1}{3} \). This gives \( \frac{2}{3} \lim _{x \rightarrow 0} \frac{\sin^2(x)}{x} \).

Step 5: Use Basic Limit Property

Use the property that \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \). Thus, \( \lim_{x \to 0} \left( \frac{\sin(x)}{x} \right)^2 = 1^2 = 1 \).

Step 6: Evaluate the Final Limit Expression

Multiply the evaluated limit: \( \frac{2}{3} \cdot 1 = \frac{2}{3} \). Thus, the original limit is \( \frac{2}{3} \).

Key Concepts

Trigonometric Identity

Trigonometric identities are powerful tools that help simplify expressions involving trigonometric functions. In this exercise, we specifically utilized the identity:

• \(1 - \cos(\theta) = 2 \sin^2(\theta/2)\)

This particular identity is crucial because it allows us to express the difference between 1 and the cosine of a multiple angle in terms of sine. In our problem, the angle \(\theta\) is equivalent to \(2x\), leading to the substitution \(1 - \cos(2x) = 2 \sin^2(x)\).
This step creates a pathway for transforming complex trigonometric expressions into a more manageable form, often allowing us to exploit known limits or simplify derivatives. Remember, mastering these identities not only simplifies calculations but also deepens understanding of trigonometric behaviors.

Limit Evaluation

Limit evaluation is a fundamental concept in calculus. It involves determining the value that a function approaches as the input approaches a certain point. In this exercise, we need to evaluate the limit as \(x\) approaches 0 for the expression \(\lim _{x \rightarrow 0} \frac{1-\cos(2x)}{3x}\).
By applying the trigonometric identity, we transform the original expression into \(\lim _{x \rightarrow 0} \frac{2 \sin^2(x)}{3x}\). This strategic replacement helps us simplify the limit expression.
To evaluate the limit, we first rearrange it to \(\frac{2}{3} \lim _{x \rightarrow 0} \frac{\sin^2(x)}{x}\). Now the task reduces to solving this simpler form. Knowing which identities and properties to apply is key to successful limit evaluation.

Basic Limit Property

When simplifying limits involving trigonometric functions, basic limit properties often become indispensable. One of the most valuable properties used in trigonometric limits is:

• \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\)

Understanding this property is crucial because it provides an easy way to handle limits involving sine functions. In our case, the expression became \(\frac{2}{3} \lim _{x \rightarrow 0} \left( \frac{\sin(x)}{x} \right)^2\).
By using the basic limit property, \(\left( \frac{\sin(x)}{x} \right)^2\) approaches 1 as \(x\) approaches 0. This simplification allows us to compute \(\frac{2}{3} \times 1 = \frac{2}{3}\). Mastering such basic limit properties can significantly speed up resolving complex-looking limits.