Problem 65 - 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. $$\lim _{(x, y) \rightarrow(0,1)} \frac{y^{2} \sin x}{x}$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understand the Function

We need to evaluate the limit \( \lim _{(x, y) \rightarrow(0,1)} \frac{y^{2} \sin x}{x} \). The function is dependent on \(x\) and \(y\), where the limit is approaching the point \((0, 1)\). Specifically, \(y\) tends to 1 and \(x\) tends to 0.

Step 2: Substitute \(y\) Approaching 1

Since \(y\) approaches 1 as \((x, y) \to (0, 1)\), substitute \(y=1\) into the function. The expression becomes \( \frac{1^{2} \sin x}{x} = \frac{\sin x}{x} \).

Step 3: Evaluate the Standard Limit \( \lim _{x \to 0} \frac{\sin x}{x} \)

It is known that \( \lim _{x \to 0} \frac{\sin x}{x} = 1 \). This is a standard limit in calculus, based on the behavior of the sine function around zero.

Key Concepts

Limit of a function

In calculus, the limit of a function describes the value that a function approaches as the input approaches some value. Limits are essential to understanding calculus, enabling us to define concepts like continuity and derivatives. When dealing with limits of functions of two variables, the situation gets more complex.

In this exercise, we are asked to compute the limit of a multivariable function as \((x, y)\) approaches the point \((0, 1)\). Here, it means:

• As \(x\) approaches 0
• As \(y\) approaches 1

Given that the function is \(\frac{y^{2} \sin x}{x}\), we need to substitute the value \(y = 1\). This results in a single-variable function \(\frac{\sin x}{x}\) as \(x\) approaches 0. Evaluating such limits involves understanding the behavior of the function very close to the point we are interested in. If a consistent value can be approached from any direction, then the limit exists.

For this problem, our function simplifies to a standard limit form, known as \( \lim _{x \to 0} \frac{\sin x}{x} = 1\), which is especially significant in the study of calculus.

Sine function

The sine function \(\sin x\) is a fundamental concept in trigonometry and calculus. It represents one of the primary functions used to describe oscillatory behavior such as waves. The function cycles between -1 and 1, repeatedly in the interval of \(\left[0, 2\pi\right]\).

The form \(\frac{\sin x}{x}\) arises frequently in calculus because it describes the ratio between the sine of an angle and the angle itself, measured in radians. As \(x\) approaches zero, the value of \(\frac{\sin x}{x}\) approaches 1. Understanding this behavior is crucial because it forms the basis for many approximations and further calculus concepts like Taylor series and differentials.

Whenever a limit involves sine, such as \(\lim _{x \to 0} \frac{\sin x}{x}\), we recognize that this limit equals 1, due to its geometric interpretation or by using the small angle theorem. This extraordinary characteristic allows us to simplify multivariable problems by reducing them to known single-variable limits.

Approaching a point in multivariable functions

In multivariable calculus, approaching a point doesn't just involve moving along the x-axis or y-axis as it often does in single-variable calculus. You must consider simultaneous changes in both axes, which makes evaluating limits a bit more challenging when working with multivariable functions.

Multivariable functions can approach a point from multiple paths and directions. For the limit to exist at a given point, the function should approach the same value no matter the path taken towards that point. This is different from single-variable calculus where you only consider left-hand and right-hand limits.

• Different paths to the same point may give different limit values, indicating that the overall limit at that point does not exist.
• A common approach to ascertain the existence of a limit is to simplify the function to a lower-dimensional situation, as done in this exercise by substituting \(y = 1\) and analyzing \(\sin x/x\) as \(x\) approaches zero.

Understanding how to navigate these concepts is fundamental in mastering multivariable calculus, and allows students to tackle more complex problems with confidence.