Problem 74 - Calculus Volume 3

Question

For the following exercises, evaluate the limits at the indicated values of \(x\) and \(y\). If the limit does not exist, state this and explain why the limit does not exist. $$\lim _{(x, y) \rightarrow(\pi, \pi)} x \sin \left(\frac{x+y}{4}\right)$$

Step-by-Step Solution

Answer

The limit is \(\pi\).

Step 1: Understanding Limit Expression

The expression we need to evaluate is \( \lim _{(x, y) \rightarrow(\pi, \pi)} x \sin \left(\frac{x+y}{4}\right) \). This asks us to find the value the expression approaches as \((x, y)\) gets arbitrarily close to \((\pi, \pi)\).

Step 2: Substituting the Limit Point

We start by substituting \(x = \pi\) and \(y = \pi\) into the expression to evaluate the limit directly. This gives us \( \pi \sin \left(\frac{\pi+\pi}{4}\right) \).

Step 3: Simplifying Inside the Sine Function

Calculate \(\frac{\pi + \pi}{4}\) which simplifies to \(\frac{2\pi}{4} = \frac{\pi}{2}\).

Step 4: Evaluating the Sine Function

Next, we find \(\sin\left(\frac{\pi}{2}\right)\), which is known to be \(1\).

Step 5: Calculating the Final Limit Value

Substitute the sine value back into the expression: \( \pi \times 1 = \pi \). Thus, the limit evaluates to \( \pi \).

Key Concepts

Evaluating Limits

In multivariable calculus, evaluating limits involves finding the value that a function approaches as the variables within the function get close to a specified point. Here, we are interested in the point \((x, y) \to (\pi, \pi)\). This means both \(x\) and \(y\) approach \(\pi\). Evaluating limits can be more complex in multivariable calculus compared to single-variable calculus, as it considers various paths along which the variables can approach the point.

• The concept of a limit in this context is about the behavior of a function as it nears a particular point.
• In this problem, we are examining how the expression behaves as \((x, y)\) nears \((\pi, \pi)\) specifically.
• Tools like algebraic simplification or substitution are often used to evaluate such limits.

By understanding the general approach to the limit point, students can better grasp how particular paths or substitutions impact the limit's evaluation.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, play a crucial role in calculus. In this exercise, we are dealing with the sine function — often represented in formulas as \( \sin \). These functions are not just periodic but also fundamentally linked with geometry and waves. Understanding these functions is key when dealing with limits that include trigonometric terms.

• Trigonometric functions are functions that relate to the angles and sides of right-angled triangles.
• They are periodic, meaning they have repeating cycles. For sine, this period is \(2\pi\).
• These functions are crucial in physics, engineering, and many fields involving waves and oscillations.

In our problem, the sine function helps determine how the product of \(x\) and \(y\) changes as they approach \(\pi\).

Direct Substitution

Direct substitution is one of the simplest methods to evaluate limits. It involves directly replacing the variables in the function with their approaching values and then simplifying the result.

• If the function is continuous and straightforward around the limit point, this method is very efficient.
• The goal is to substitute the coordinates that \(x\) and \(y\) approach directly into the given function.
• In some instances, direct substitution can lead to indeterminate forms, requiring additional strategies for resolution.

In our exercise, direct substitution helped us quickly find the value of the expression at \(x = \pi\) and \(y = \pi\). These straightforward cases allow us to solve the limit without additional effort beyond basic arithmetic and understanding of sine values.

Sine Function

The sine function, denoted as \(\sin\), is a basic trigonometric function that returns the sine of a given angle. It's fundamental in periodic and wave phenomena, often cycling from -1 to 1. When considering limits, the sine function's behavior at specific points is well-documented, aiding easy evaluation.

• The sine function peaks at \(1\) and \(-1\); it is zero at \(0\), \(\pi\), and its multiples.
• In the context of our exercise, the evaluation of \(\sin\left(\frac{\pi}{2}\right) = 1\) is a critical step.
• This specific angle \(\frac{\pi}{2}\) is known as the maximum point in the sine curve, making the evaluation straightforward.

By knowing such key values of the sine function, you simplify the calculation process when evaluating limits or tackling related calculus problems.