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Chapter 12: Problem 72

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

Short Answer

Expert verified
In summary, both given limits have the value 1 as (x, y) approaches (0, 0). For part a, we used the variable substitution to evaluate the limit, and for part b, we decomposed the expression into known limit forms and evaluated each term separately.

Step by step solution

01

Use substitute variable t

Let \(t = x + y\). Then, as \((x, y) \rightarrow (0,0)\), \(t \rightarrow 0\). This allows us to re-write our limit as: \(\lim _{t\to 0} \frac{\sin t}{t}\)
02

Evaluate the limit

We know that the limit \(\lim_{t\to 0} \frac{\sin t}{t}\) is equal to 1. Therefore, the limit of our original expression is also 1: \(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin (x+y)}{x+y} = 1\) b. \(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin x+\sin y}{x+y}\)
03

Break the fraction into two fractions

We can rewrite the given expression as the sum of two fractions: \(\frac{\sin x+\sin y}{x+y}= \frac{\sin x}{x+y}+\frac{\sin y}{x+y}\)
04

Add and subtract a term

To evaluate the limit, we can add and subtract a term in each fraction, allowing us to use limit properties: \(\frac{\sin x}{x+y}+\frac{\sin y}{x+y} = \frac{\sin x - x + x}{x+y} + \frac{\sin y - y + y}{x+y}\)
05

Evaluate the limit

Now we can break up each fraction further and use limit properties: \(\lim_{(x, y)\to (0,0)}(\frac{\sin x - x}{x+y} + \frac{x}{x+y} + \frac{\sin y - y}{x+y} + \frac{y}{x+y})\) Using the fact that the limit of a sum is the sum of the limits and the limits \(\lim_{t \to 0} \frac{\sin t - t}{t} = \lim_{t \to 0}\frac{t(\cos t - 1)}{t} = 0\) and \(\lim_{t\to 0} \frac{\sin t}{t} = 1\), we can evaluate each term: \(\lim_{(x, y)\to (0,0)}\frac{\sin x - x}{x+y} = 0\) \(\lim_{(x, y)\to (0,0)}\frac{x}{x+y} = \frac{1}{2}\) \(\lim_{(x, y)\to (0,0)}\frac{\sin y - y}{x+y} = 0\) \(\lim_{(x, y)\to (0,0)}\frac{y}{x+y} = \frac{1}{2}\) Summing these limits together, we get the final answer: \(\lim _{(x, y) \rightarrow(0,0)} \frac{\sin x+\sin y}{x+y} = 0 + \frac{1}{2} + 0+ \frac{1}{2} = 1\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Evaluation
Limit evaluation is a fundamental skill in calculus, particularly when dealing with multivariable functions. It involves finding the value that a function approaches as its input variables approach specific points. This concept requires a solid understanding of limits from single-variable calculus and extends it into more dimensions.

When evaluating multivariable limits, such as \( \lim _{(x, y) \rightarrow(0,0)} \frac{\sin (x+y)}{x+y} \), you often look for strategies to simplify the function. The key is to find a path to zero that makes the limit straightforward to compute. Sometimes, direct substitution isn't possible, and other techniques are needed.

Evaluating limits can become intricate, particularly with expressions involving quotients. The substitution method, as seen in these exercises, becomes particularly handy as it can convert a more complex multivariable limit into a familiar single-variable limit. Successful limit evaluation can confirm the function鈥檚 behavior near a point, which is critical for understanding continuity and behavior around that area.
Trigonometric Limits
Trigonometric limits, such as \( \lim_{t\to 0} \frac{\sin t}{t} = 1 \), play a significant role in calculus. These limits often appear in problems involving trigonometric functions and are crucial for evaluating the behavior of these functions near certain points.

The limit \( \lim_{t\to 0} \frac{\sin t}{t} = 1 \) is a fundamental result used frequently when simplifying expressions. It states that, as \( t \) gets very close to zero, the ratio of \( \sin t \) to \( t \) approaches 1.
  • This limit is pivotal in solving exercises involving small angle approximations, where \( \sin t \approx t \) for small \( t \).
  • It also helps in constructing other limits that involve composing functions of trigonometric terms.
Understanding trigonometric limits helps bridge the gap between algebraic manipulation and intuitive geometric understanding. These limits often require built-in knowledge of certain trigonometric identities and their behaviors around zero.
Substitution Method
The substitution method is a technique used for simplifying limit problems by introducing new variables to make the problem more recognizable or easier to handle. This is particularly useful for multivariable calculus problems. It involves making a substitution to transform the original limit into one that is simpler or already known.

For instance, in the problem \( \lim _{(x, y) \rightarrow(0,0)} \frac{\sin (x+y)}{x+y} \), the substitution \( t = x + y \) reduces the two-variable problem to a single variable \( t \) as it approaches zero. This allows us to use existing knowledge about single-variable limits.
  • By substituting, we convert a complex function into a more manageable form.
  • This method is also crucial for leveraging known limits, like \( \lim_{t\to 0} \frac{\sin t}{t} = 1 \), to simplify calculations.
The substitution method is an effective way to tackle complicated forms, enabling easier manipulation and solution of difficult calculus problems. It emphasizes the importance of looking at problems from different angles to find the simplest path to a solution.

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Most popular questions from this chapter

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Let the equation of the best-fit line be \(y=m x+b,\) where the slope \(m\) and the \(y\) -intercept \(b\) must be determined using the least squares condition. First assume that there are three data points \((1,2),(3,5),\) and \((4,6) .\) Show that the function of \(m\) and \(b\) that gives the sum of the squares of the vertical distances between the line and the three data points is $$ \begin{aligned} E(m, b)=&((m+b)-2)^{2}+((3 m+b)-5)^{2} \\ &+((4 m+b)-6)^{2} \end{aligned}. $$ Find the critical points of \(E\) and find the values of \(m\) and \(b\) that minimize \(E\). Graph the three data points and the best-fit line.

Let \(w=f(x, y, z)=2 x+3 y+4 z\) which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\) \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\)

Identify and briefly describe the surfaces defined by the following equations. $$z^{2}+4 y^{2}-x^{2}=1$$

a. Show that the point in the plane \(a x+b y+c z=d\) nearest the origin is \(P\left(a d / D^{2}, b d / D^{2}, c d / D^{2}\right),\) where \(D^{2}=a^{2}+b^{2}+c^{2} .\) Conclude that the least distance from the plane to the origin is \(|d| / D\). (Hint: The least distance is along a normal to the plane.) b. Show that the least distance from the point \(P_{0}\left(x_{0}, y_{0}, z_{0}\right)\) to the plane \(a x+b y+c z=d\) is \(\left|a x_{0}+b y_{0}+c z_{0}-d\right| / D\) (Hint: Find the point \(P\) on the plane closest to \(P_{0}\).)

Recall that Cartesian and polar coordinates are related through the transformation equations $$\left\\{\begin{array}{l} x=r \cos \theta \\ y=r \sin \theta \end{array} \quad \text { or } \quad\left\\{\begin{array}{l} r^{2}=x^{2}+y^{2} \\ \tan \theta=y / x \end{array}\right.\right.$$ a. Evaluate the partial derivatives \(x_{r}, y_{r}, x_{\theta},\) and \(y_{\theta}\) b. Evaluate the partial derivatives \(r_{x}, r_{y}, \theta_{x},\) and \(\theta_{y}\) c. For a function \(z=f(x, y),\) find \(z_{r}\) and \(z_{\theta},\) where \(x\) and \(y\) are expressed in terms of \(r\) and \(\theta\) d. For a function \(z=g(r, \theta),\) find \(z_{x}\) and \(z_{y},\) where \(r\) and \(\theta\) are expressed in terms of \(x\) and \(y\) e. Show that \(\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}=\left(\frac{\partial z}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial z}{\partial \theta}\right)^{2}\)
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