Problem 8 - Calculus Volume 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 is \( \langle 1, \frac{1}{4}, 1 \rangle \).

Step 1: Analyzing the first component

For the vector function \( \langle \sqrt{t-3}, \frac{\sqrt{t}-2}{t-4}, \tan \left(\frac{\pi}{t}\right) \rangle \), let's first consider the limit of the first component \( \sqrt{t-3} \) as \( t \rightarrow 4 \). Substitute \( t = 4 \) to get \( \sqrt{4-3} = \sqrt{1} = 1 \).

Step 2: Simplifying the second component

Now consider the second component \( \frac{\sqrt{t}-2}{t-4} \). This becomes an indeterminate form \( \frac{0}{0} \) as \( t \to 4 \). To solve this, apply L'Hôpital's rule which requires differentiating the numerator and denominator separately.The derivative of the numerator \( \sqrt{t} \) is \( \frac{1}{2\sqrt{t}} \) and the derivative of the denominator \( t - 4 \) is \( 1 \). Thus, the limit becomes:\[ \lim_{t \to 4} \frac{1}{2\sqrt{t}} = \frac{1}{2\sqrt{4}} = \frac{1}{4}. \]

Step 3: Evaluating the third component

For the third component \( \tan \left( \frac{\pi}{t} \right) \), substitute \( t = 4 \) directly to get \( \tan \left( \frac{\pi}{4} \right) \). We know \( \tan \left( \frac{\pi}{4} \right) = 1 \).

Step 4: Combining all components

Now, combine the limits of all components to get:\[ \lim_{t \rightarrow 4} \langle \sqrt{t-3}, \frac{\sqrt{t}-2}{t-4}, \tan \left( \frac{\pi}{t} \right) \rangle = \langle 1, \frac{1}{4}, 1 \rangle. \]

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus that helps find limits of indeterminate forms. Indeterminate forms are situations where substituting a limit value into a function results in an undefined expression like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). In the second component of the vector-valued function we analyzed, \( \frac{\sqrt{t}-2}{t-4} \), substituting \( t = 4 \) leads to the form \( \frac{0}{0} \). This is where L'Hôpital's Rule comes in handy.

According to L'Hôpital's Rule:

• Differential the numerator and the denominator separately.
• In our case, the derivative of \( \sqrt{t} \) is \( \frac{1}{2\sqrt{t}} \) and the derivative of \( t-4 \) is \( 1 \).
• Plug these derivatives back into the limit \( \lim_{t \to 4} \frac{1}{2\sqrt{t}} \).
• You can now find the limit as \( t \to 4 \) without the indeterminate form, which simplifies to \( \frac{1}{4} \).

This technique is essential for dealing with fractions that seem unsolvable by basic substitution. It makes otherwise challenging limits approachable.

Indeterminate Forms

Indeterminate forms occur when evaluating limits leads to uncertain or undefined expressions, typically \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms do not tell us anything conclusive about the behavior of a function near a specific point. Because of this uncertainty, they require special methods to evaluate.

When dealing with indeterminate forms like \( \frac{\sqrt{t}-2}{t-4} \), straightforward substitution of \( t = 4 \) gives \( \frac{0}{0} \), an indeterminate form. This calls for more advanced techniques like:

• Simplifying the expression - Sometimes multiplying by a conjugate can help eliminate indeterminate forms.
• L'Hôpital's Rule - As mentioned above, is specifically designed for when you encounter \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Algebraic manipulation - Rearranging and factoring can occasionally resolve the indeterminate form without any derivative calculations.

Recognizing the indeterminate form is the first step, and choosing the right strategy is crucial for successfully finding the limit.

Trigonometric Limits

Trigonometric limits involve understanding how trigonometric functions like sine, cosine, and tangent behave as their input approaches certain critical values. These limits often appear in problems involving oscillatory behavior or periodic functions. In the given exercise, the third component of the function, \( \tan \left( \frac{\pi}{t} \right) \), is a trigonometric limit we must evaluate.

For trigonometric limits, substituting the limit value directly is often a good start. For example:

• By substituting \( t = 4 \) in \( \tan \left( \frac{\pi}{t} \right) \), it simplifies to \( \tan \left( \frac{\pi}{4} \right) \).
• We know from basic trigonometry that \( \tan \left( \frac{\pi}{4} \right) = 1 \). There are no complications here.

Understanding basic trigonometric values can drastically simplify the process. Sometimes, especially with more complex limits, using trigonometric identities or limit properties might be needed. Fortunately, in our example, a simple substitution was all it took to find the value of the limit.