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Chapter 13: Problem 26

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

Short Answer

Expert verified
Question: Determine the value of the following limit: $$\lim _{(x, y) \rightarrow(1,2)} \frac{\sqrt{y}-\sqrt{x+1}}{y-x-1}$$ Answer: The value of the given limit is $$\frac{\sqrt{2}}{4}.$$

Step by step solution

01

Analyzing the given limit

Given the limit: $$\lim _{(x, y) \rightarrow(1,2)} \frac{\sqrt{y}-\sqrt{x+1}}{y-x-1}$$ We want to find the limit of the function as \((x,y) \rightarrow (1,2)\). Let's try direct substitution:
02

Direct substitution

Substitute \((x,y)=(1,2)\) into the function: $$\frac{\sqrt{2}-\sqrt{1+1}}{2-1-1} = \frac{\sqrt{2}-\sqrt{2}}{0}$$ We can see that we have a form \(\frac{0}{0}\), which is indeterminate. Direct substitution didn't work. Now, let's try an algebraic method called Rationalization.
03

Rationalization

First, let's rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator: $$\frac{\sqrt{y}-\sqrt{x+1}}{y-x-1} * \frac{\sqrt{y}+\sqrt{x+1}}{\sqrt{y}+\sqrt{x+1}} = \frac{y-(x+1)}{(y-x-1)(\sqrt{y}+\sqrt{x+1})}$$ Now, simplify the expression:
04

Simplify the expression

Simplify the expression obtained after rationalization: $$\frac{y-x-1}{(y-x-1)(\sqrt{y}+\sqrt{x+1})} = \frac{1}{\sqrt{y}+\sqrt{x+1}}$$ At this point, we have a simplified expression that we should now be able to evaluate the limit for.
05

Direct substitution (again)

Substitute \((x,y)=(1,2)\) into the simplified expression: $$\lim _{(x, y) \rightarrow(1,2)} \frac{1}{\sqrt{y}+\sqrt{x+1}} = \frac{1}{\sqrt{2}+\sqrt{1+1}} = \frac{1}{\sqrt{2}+\sqrt{2}} = \frac{1}{2\sqrt{2}}$$ Now, simplify the denominator to obtain the final answer:
06

Final simplification

To further simplify, we can multiply the numerator and denominator by \(\sqrt{2}\): $$\frac{1}{2\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{4}$$ So, the limit is equal to the final result: $$\lim _{(x, y) \rightarrow(1,2)} \frac{\sqrt{y}-\sqrt{x+1}}{y-x-1} = \frac{\sqrt{2}}{4}$$

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Most popular questions from this chapter

In the advanced subject of complex variables, a function typically has the form \(f(x, y)=u(x, y)+i v(x, y),\) where \(u\) and \(v\) are real-valued functions and \(i=\sqrt{-1}\) is the imaginary unit. A function \(f=u+i v\) is said to be analytic (analogous to differentiable) if it satisfies the Cauchy-Riemann equations: \(u_{x}=v_{y}\) and \(u_{y}=-v_{x}\) a. Show that \(f(x, y)=\left(x^{2}-y^{2}\right)+i(2 x y)\) is analytic. b. Show that \(f(x, y)=x\left(x^{2}-3 y^{2}\right)+i y\left(3 x^{2}-y^{2}\right)\) is analytic. c. Show that if \(f=u+i v\) is analytic, then \(u_{x x}+u_{y y}=0\) and \(v_{x x}+v_{y y}=0\)

Let \(R\) be a closed bounded set in \(\mathbb{R}^{2}\) and let \(f(x, y)=a x+b y+c,\) where \(a, b,\) and \(c\) are real numbers, with \(a\) and \(b\) not both zero. Give a geometrical argument explaining why the absolute maximum and minimum values of \(f\) over \(R\) occur on the boundaries of \(R\)

Use the gradient rules of Exercise 81 to find the gradient of the following functions. $$f(x, y, z)=\sqrt{25-x^{2}-y^{2}-z^{2}}$$

Consider the curve \(\mathbf{r}(t)=\langle\cos t, \sin t, c \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(c\) is a real number. a. What is the equation of the plane \(P\) in which the curve lies? b. What is the angle between \(P\) and the \(x y\) -plane? c. Prove that the curve is an ellipse in \(P\).

Use the gradient rules of Exercise 81 to find the gradient of the following functions. $$f(x, y, z)=(x+y+z) e^{x y z}$$
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