Problem 68 - Precalculus with Limits

Question

In Exercises 49-68, find the limit by direct substitution. $$ \lim_{x \to 1}\ \textrm{arccos}\ \dfrac{x}{2}$$

Step-by-Step Solution

Answer

Thus, the limit of the function as \(x\) approaches 1 is \(π/3\) or \(60°\).

Step 1: Direct Substitution

Substitute \(x = 1\) into the function, i.e., replace \(x\) with 1 in \(\textrm{arccos}\ \dfrac{x}{2}\).

Step 2: Calculation of arccosine

After the substitution, the expression becomes \(\textrm{arccos}\ \dfrac{1}{2}\). This yields an angle \(θ\) such that \(\cos(θ) = \dfrac{1}{2}\). The value of \(\textrm{arccos}\ \dfrac{1}{2}\) is \(π/3\) or \(60°\) in radian measure.

Key Concepts

Direct Substitution

Direct substitution is one of the simplest techniques used to find limits in calculus. It involves directly plugging the value approaching the variable into the function. If the function is well-behaved and continuous at the point you're considering, this method works like a charm. This is often the first step in evaluating limits because:

• It's straightforward and direct.
• Often provides immediate insight into the behavior of the function as the variable approaches a particular value.
• Can simplify the process of understanding more complex limit problems.

For example, with the expression \( \lim_{x \to 1}\ \textrm{arccos}\ \dfrac{x}{2} \), we substitute \( x = 1 \) directly into the function. By doing so, we calculate \( \textrm{arccos}\ \frac{1}{2} \), avoiding any complications from variables or indeterminate forms. If you start with this technique and find it doesn't provide an immediate answer, it might indicate the function isn't continuous or differentiable at that point, leading you to try other techniques.

Arccosine Function

The arccosine function, denoted as \( \textrm{arccos}(x) \), is the inverse function of the cosine function over its principal range. It's vital to understanding how the function translates a value from the cosine range back to the angle. Here are some key points:

• While the cosine function takes an angle and provides a ratio (commonly defined from -1 to 1), the arccosine reverses this process.
• \( \textrm{arccos}(x) \) provides an angle between \( 0 \) and \( \pi \) where \( \cos(\theta) = x \).
• It means if you know \( \cos(\theta) = \frac{1}{2} \), then \( \theta \) is one of the angles where this is true, specifically \( \pi/3 \) radians or \( 60^{\circ} \).

The arccosine function is handy in various fields such as trigonometry and geometry, where converting a cosine value back to an angle is required. It allows for the resolution of unknown angles based on known trigonometric ratios, essential for solving workouts involving circular dimensions or periodic phenomena.

Trigonometric Limits

Trigonometric limits involve finding the limit of expressions with trigonometric functions. These limits are important in calculus since they help in understanding the behavior of trigonometric functions as the variable approaches a particular point. Here's what to keep in mind:

• They often involve functions such as \( \sin(x), \cos(x), \tan(x) \) and their inverses (like \( \arccos(x) \)).
• Knowing specific trigonometric values by heart can simplify limit calculations, like \( \lim_{x \to 0} \sin(x)/x = 1 \).
• Simplification and identities play significant roles. It’s all about recognizing and using identities like \( \sin^2(x) + \cos^2(x) = 1 \).
• For our case with \( \lim_{x \to 1} \textrm{arccos} \frac{x}{2} \), it's a straight application of substitution and trigonometric identity understanding to get \( \frac{\pi}{3} \).

Trigonometric limits can sometimes look intimidating, but by knowing key values and identities, the process becomes much smoother. Learning these limits fosters a better understanding of the cyclic nature of trigonometric functions and how they operate over different intervals.