Problem 72 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$a.\lim _{(x, y) \rightarrow(0,0)} \frac{\sin (x+y)}{x+y}$$ $$b.\lim _{(x, y) \rightarrow(0,0)} \frac{\sin x+\sin y}{x+y}$$

Step-by-Step Solution

Answer

Question: Evaluate the following limits as (x, y) approaches (0, 0): a) \(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin (x+y)}{x+y}\) b) \(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin x+\sin y}{x+y}\) Answer: a) The limit for part a is: \(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin (x+y)}{x+y} = 1\) b) The limit for part b is: \(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin x+\sin y}{x+y} = \frac{u}{2v}\)

Step 1: Combine expressions

Since both terms are combined in the limit, we can treat the limit as a single-variable limit by substituting \((x+y)\) with \(z\). So, we can change the limit to: $$\lim _{z \rightarrow 0} \frac{\sin z}{z}$$

Step 2: Known special limit

Now we have a known special limit. The single-variable limit of \(\frac{\sin(z)}{z}\) as z approaches 0 is equal to 1. So the limit for part a is: $$\lim _{(x, y) \rightarrow(0,0)} \frac{\sin (x+y)}{x+y} = 1$$ #b. Evaluate the limit \(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin x+\sin y}{x+y}\)#

Step 1: Divide the expression into separate limits

We can rewrite the expression as: $$\frac{1}{2}\lim _{(x, y) \rightarrow(0,0)} \frac{\sin x}{(x+y)/2} + \frac{1}{2}\lim _{(x, y) \rightarrow(0,0)} \frac{\sin y}{(x+y)/2}$$

Step 2: Convert to single-variable limits

Now we make substitutions to rewrite both limits into a single-variable limit. We can let \(u = x+y\), \(v = \frac{1}{2}(x+y)\), and use the fact that \(\lim_{(x,y) \to (0,0)} u = \lim_{(x,y) \to (0,0)} v = 0\). So we get: $$\frac{1}{2}\lim _{u \rightarrow 0} \frac{\sin u}{2v} + \frac{1}{2}\lim _{u \rightarrow 0} \frac{\sin u}{2v}$$ Since the limits on the right-hand side are equal, we can write: $$= \lim _{u \rightarrow 0} \frac{\sin u}{2v}$$

Step 3: Known special limit

Now we have a limit that is similar to the known special limit. We can split the limit to get: $$\lim _{u \rightarrow 0} \frac{\sin u}{u} \cdot \frac{u}{2v} = 1 \cdot \frac{u}{2v}$$ So the limit for part b is: $$\lim _{(x, y) \rightarrow(0,0)} \frac{\sin x+\sin y}{x+y} = \frac{u}{2v}$$

Key Concepts

Limits

Understanding limits is crucial in calculus as they help us comprehend the behavior of functions at points of interest. In simple terms, a limit describes what value a function approaches as the input values get closer to a certain point. Mathematically, when we write \( \lim_{x \to a} f(x) = L \) , we mean that as \( x \) gets closer and closer to \( a \), the function \( f(x) \) gets closer and closer to \( L \).

• This concept is particularly important in situations where the function may not be explicitly defined at a particular point.
• Limits can be evaluated for functions of one or multiple variables.
• When dealing with multivariable functions, we're interested in the behavior as several variables approach particular values.

In the exercise, we are dealing with a two-variable function and want to figure out what happens as both \( x \) and \( y \) approach 0. The challenge with multivariable limits is ensuring the function approaches a particular value from all possible paths. This is where techniques like substitution are handy, allowing us to simplify complex multivariable problems into more familiar single-variable ones.

Special Limits

Some limits have well-known solutions, referred to as 'special limits'. A classic example is \( \lim_{z \to 0} \frac{\sin z}{z} = 1 \). Why is this special? Let's break it down:

• This limit is fundamental because it forms the basis for many calculus identities and is often used as a rule of thumb when analyzing trigonometric limits.
• Special limits act as shortcuts, making it quicker and simpler to compute complex expressions by recognizing and applying known results.
• Instead of going through lengthy proofs or calculations, we can rely on these known limits to deduce the outcome efficiently.

In the exercise's first part, recognizing that \( \frac{\sin(x+y)}{x+y} \) aligns with this special limit simplifies our work. By treating \( x+y \) as a single entity, it's straightforward to see why this expression approaches 1 when both variables head to zero.

Single-Variable Substitution

Substitution is a clever tool in calculus, simplifying complex multivariable limits by reducing them to more familiar single-variable forms. Here's how it works:

• We identify common terms or expressions that can be represented by a single variable.
• By substituting, we translate a multi-dimensional problem into a simpler, one-dimensional problem.
• It allows for the application of known single-variable limits and techniques.

For example, in the given exercise, instead of handling two variables \( x \) and \( y \) separately, we can combine them into one variable \( z = x+y \), making it easier to evaluate the limit. In part b of the exercise, expressions are further manipulated using substitutions like \( u \) and \( v \) to break down a more intricate limit into components that match known limit forms. This approach streamlines solving by focusing on one variable at a time, applying known solutions, and simplifying tasks like edge behavior analysis and continuity assessment.