/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Evaluate the following limits. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 25

Evaluate the following limits. $$\lim _{(x, y) \rightarrow(1,2)} \frac{\sqrt{y}-\sqrt{x+1}}{y-x-1}$$

Short Answer

Expert verified
Question: Evaluate the limit of the expression as (x, y) approaches (1, 2): $$\lim _{(x, y) \rightarrow(1,2)} \frac{\sqrt{y}-\sqrt{x+1}}{y-x-1}$$ Answer: The limit of the given expression as (x, y) approaches (1, 2) is $$\frac{1}{2\sqrt{2}}$$.

Step by step solution

01

Observe the given limit expression

Observe the given limit expression: $$\lim _{(x, y) \rightarrow(1,2)} \frac{\sqrt{y}-\sqrt{x+1}}{y-x-1}$$
02

Perform a change of variables

To simplify the expression, perform a change of variables: let \(u = x + 1\) and \(v = y\). Now, as \((x, y) \rightarrow (1, 2)\), we have \((u, v) \rightarrow (2, 2)\). Rewrite the limit expression using \(u\) and \(v\): $$\lim _{(u, v) \rightarrow(2, 2)} \frac{\sqrt{v}-\sqrt{u}}{v-u}$$
03

Apply the conjugate to the numerator

To further simplify the expression, multiply the numerator and denominator by the conjugate of the numerator, which is \(\sqrt{v} + \sqrt{u}\). This can help to remove square roots from the expression: $$\lim _{(u, v) \rightarrow(2, 2)} \frac{(\sqrt{v}-\sqrt{u})(\sqrt{v}+\sqrt{u})}{(v-u)(\sqrt{v}+\sqrt{u})}$$
04

Simplify the expression

Now, simplify the expression by calculating the product of the numerators and updating the denominators: $$\lim _{(u, v) \rightarrow(2, 2)} \frac{v-u}{(v-u)(\sqrt{v}+\sqrt{u})}$$
05

Simplify the expression further

The expression can be further simplified as the \((v-u)\) term cancels out from the numerator and denominator: $$\lim _{(u, v) \rightarrow(2, 2)} \frac{1}{\sqrt{v}+\sqrt{u}}$$
06

Substitute the limit values

Now that the expression is simplified, we can substitute the values for \(u\) and \(v\) as they approach \((2, 2)\) to evaluate the limit: $$\lim _{(u, v) \rightarrow(2, 2)} \frac{1}{\sqrt{v}+\sqrt{u}} = \frac{1}{\sqrt{2}+\sqrt{2}} = \frac{1}{2\sqrt{2}}$$ So, the limit of the given expression as \((x, y) \rightarrow (1, 2)\) is \(\frac{1}{2\sqrt{2}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use the formal definition of a limit to prove that $$\lim _{(x, y) \rightarrow(a, b)} c f(x, y)=c \lim _{(x, y) \rightarrow(a, b)} f(x, y)$$

Match equations a-f with surfaces A-F. a. \(y-z^{2}=0\) b. \(2 x+3 y-z=5\) c. \(4 x^{2}+\frac{y^{2}}{9}+z^{2}=1\) d. \(x^{2}+\frac{y^{2}}{9}-z^{2}=1\) e. \(x^{2}+\frac{y^{2}}{9}=z^{2}\) f. \(y=|x|\)

Ideal Gas Law Many gases can be modeled by the Ideal Gas Law, \(P V=n R T,\) which relates the temperature \((T,\) measured in Kelvin (K)), pressure ( \(P\), measured in Pascals (Pa)), and volume ( \(V\), measured in \(\mathrm{m}^{3}\) ) of a gas. Assume that the quantity of gas in question is \(n=1\) mole (mol). The gas constant has a value of \(R=8.3 \mathrm{m}^{3} \cdot \mathrm{Pa} / \mathrm{mol} \cdot \mathrm{K}.\) a. Consider \(T\) to be the dependent variable and plot several level curves (called isotherms) of the temperature surface in the region \(0 \leq P \leq 100,000\) and \(0 \leq V \leq 0.5.\) b. Consider \(P\) to be the dependent variable and plot several level curves (called isobars) of the pressure surface in the region \(0 \leq T \leq 900\) and \(0< V \leq 0.5.\) c. Consider \(V\) to be the dependent variable and plot several level curves of the volume surface in the region \(0 \leq T \leq 900\) and \(0 < P \leq 100,000.\)

Absolute maximum and minimum values Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}-4 y^{2}+x y ; R=\left\\{(x, y): 4 x^{2}+9 y^{2} \leq 36\right\\}$$

Problems with two constraints Given a differentiable function \(w=f(x, y, z),\) the goal is to find its maximum and minimum values subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0,\) and \(h(x, y, z)=0\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.