Problem 8 - Calculus 3

Question

Find the limit of the following vector-valued functions at the indicated value of \(t .\) $$ \lim _{t \rightarrow 4}\left\langle\sqrt{t-3}, \frac{\sqrt{t}-2}{t-4}, \tan \left(\frac{\pi}{t}\right)\right\rangle $$

Step-by-Step Solution

Answer

The limit of the vector is \( \left\langle 1, \frac{1}{4}, \tan\left(\frac{\pi}{4}\right) \right\rangle \).

Step 1: Examine Each Component

We need to find the limit of each component of the vector separately as \( t \rightarrow 4 \). We have a vector \( \left\langle \sqrt{t-3}, \frac{\sqrt{t}-2}{t-4}, \tan\left(\frac{\pi}{t}\right) \right\rangle \). The components are \( \sqrt{t-3} \), \( \frac{\sqrt{t}-2}{t-4} \), and \( \tan\left(\frac{\pi}{t}\right) \).

Step 2: Identify the relevant trigonometric identities

Based on the given expression or equation, identify which trigonometric identities (Pythagorean, double-angle, sum/difference, etc.) are applicable.

Step 3: Apply the identities and simplify

Apply the identified identities to transform the expression. Simplify step by step, combining like terms and reducing fractions where possible.

Step 4: Solve or evaluate

If solving an equation, isolate the trigonometric function and find the angle(s). If evaluating, compute the final numerical value.

Step 5: State the result

Express the final answer, including all solutions in the required domain if solving an equation.

Step 6: Conclude with the answer

The limit of the vector is \( \left\langle 1, \frac{1}{4}, \tan\left(\frac{\pi}{4}\right) \right\rangle \).

Key Concepts

Component-Wise Limit

When dealing with vector-valued functions, it's important to understand that each vector component needs to be treated individually when finding the limit. This is known as the component-wise limit approach. Instead of trying to find the limit of a vector directly, break it down into its components. In this particular exercise, we have a vector with three components: - \( \sqrt{t-3} \) - \( \frac{\sqrt{t}-2}{t-4} \) - \( \tan\left(\frac{\pi}{t}\right) \) For each component, you will calculate the limit as \( t \) approaches the given value, which in this case is 4. It’s like finding the limit for a single-variable function, but you'll do so separately for each part of the vector function. This step-by-step approach simplifies the problem and helps to manage complex components efficiently.

Limit of Square Root Function

Evaluating the limit involving a square root function can often be straightforward. For the component \( \sqrt{t-3} \), observe what happens as \( t \) approaches 4. Just plug \( t = 4 \) into the expression and compute: \[ \sqrt{4-3} = \sqrt{1} = 1 \] This component demonstrates how limits can often result in simplifying expressions by direct evaluation. It's crucial to note that the limit exists because the operation inside the square root remains valid (non-negative), and approaching the value of \( t \) doesn't lead to undefined conditions. Break down each step, just in case expressions become more complicated in other contexts.

Limit of Tangent Function

Finding the limit of the tangent function, \( \tan\left(\frac{\pi}{t}\right) \), introduces some interesting behaviors, especially as \( t \) approaches small numbers or specific critical points. However, as \( t \rightarrow 4 \), the argument of the tangent function \( \frac{\pi}{4} \) is a recognizable angle. The tangent of \( \frac{\pi}{4} \) is a fundamental trigonometric result: \[ \tan\left(\frac{\pi}{4}\right) = 1 \] This direct substitution technique works well here because there are no issues of the denominator approaching zero or other singular points within the expected limits. Be careful with tangent limits though, as they can become tricky near vertical asymptotes or approaching highlights like \( \pi/2 \), where the function tends toward infinity.

Indeterminate Form 0/0

The second component \( \frac{\sqrt{t}-2}{t-4} \) presents a classic example of the indeterminate form 0/0 as \( t \rightarrow 4 \). This scenario requires using a specific technique, often involving algebraic manipulation or L'Hospital's Rule, to resolve. By rationalizing or simplifying the expression (for instance, multiplying numerator and denominator by the conjugate), you may transform it into a solvable form: Sometimes, \( L'Hospital's Rule \) helps, which involves taking the derivative of the numerator and the denominator separately and then evaluating the limit. Be alert when you encounter this indeterminate form, as it often signals the need for more involved calculus techniques.

Approaching Singularities

"Approaching singularities" refers to points where a function reaches an undefined or infinite value. Singularities can occur in vectors as components approach problematic values or functions. In our particular exercise, as \( t \rightarrow 4 \), none of the vector components themselves inherently involve singularities—though the steps required to resolve indeterminate forms or manage trigonometric limits often involve avoiding such issues. - Always check if the function becomes undefined as it approaches a specific value. - Singular points in a function could create unexpected jumps or asymptotes, so understanding their behavior is key. Strategy in these cases often involves algebraic manipulation, substitution, or calculus techniques like derivatives. Knowing how to handle limits approaching singular points is crucial to keep your solutions valid and precise.