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Let $$f(x, y)=2 x^{2}+3 y^{2}$$ Use the \(\epsilon-\delta\) definition of limits to show that $$\lim _{(x, y) \rightarrow(0,0)} f(x, y)=0$$

Short Answer

Expert verified
The limit is 0, using the \( \epsilon-\delta \) definition of limits.

Step by step solution

01

Understanding the Limit Statement

We want to show that \( \lim_{(x, y) \rightarrow (0,0)} f(x, y) = 0 \). This means we need to prove that for any \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( \sqrt{x^2 + y^2} < \delta \), it follows that \( |f(x, y) - 0| < \epsilon \).
02

Expressing the Function in Terms of \( (x, y) \)

The function is given as \( f(x, y) = 2x^2 + 3y^2 \). We need \( |f(x, y)| = |2x^2 + 3y^2| \) to be less than \( \epsilon \).
03

Bounding \( |f(x, y)| \) with \( x^2 + y^2 \)

Observe that \( |f(x, y)| = |2x^2 + 3y^2| \leq 2x^2 + 3y^2 \leq 2(x^2 + y^2) + 3(x^2 + y^2) = 5(x^2 + y^2) \).
04

Expressing \( (x, y) \) in Polar Coordinates

Convert \( (x, y) \) to polar coordinates where \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \), thus \( x^2 + y^2 = r^2 \).
05

Establishing a Relationship Between \( \delta \) and \( \epsilon \)

Now, \( 5(x^2 + y^2) = 5r^2 \). We want \( 5r^2 < \epsilon \), hence \( r^2 < \frac{\epsilon}{5} \). If \( r = \sqrt{x^2 + y^2} < \delta \), we choose \( \delta = \sqrt{\frac{\epsilon}{5}} \).
06

Concluding the \( \epsilon-\delta \) Proof

Therefore, for any \( \epsilon > 0 \), choosing \( \delta = \sqrt{\frac{\epsilon}{5}} \) ensures that \( \sqrt{x^2 + y^2} < \delta \) implies \( 5r^2 < \epsilon \), thus \( |f(x, y)| < \epsilon \). This satisfies the \( \epsilon-\delta \) definition of the limit, proving the limit statement.

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Key Concepts

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

Epsilon-Delta Definition of Limits
The epsilon-delta definition of limits is crucial in understanding continuity and the behavior of functions at specific points. It mathematically formalizes the intuitive idea of a limit. For a multivariable function like \( f(x, y) \), this definition extends from the single-variable case to two variables. In this setting, our goal is to show that as \((x, y)\) approaches \((0, 0)\), the function \(f(x, y)\) approaches a particular value, typically zero.

Formally, for \( \lim_{(x, y) \to (a, b)} f(x, y) = L \), it means for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( \sqrt{(x-a)^2 + (y-b)^2} < \delta \), then \( |f(x, y) - L| < \epsilon \). For our specific function \( f(x, y) = 2x^2 + 3y^2 \), this translates into proving the closeness of \( f(x, y) \) to zero whenever \( (x, y) \) is sufficiently close to \( (0, 0) \) within a certain radius.

This approach gives an exact method to evaluate limits and is fundamental for ensuring a function's closeness to a particular value at a given point.
Limits in Two Variables
Limits in two variables can be more complicated than in single-variable calculus because they require consideration of the point's approach direction. As \((x, y)\) approaches a target point, \((0, 0)\) in this problem, it can do so from infinitely many directions. This attribute makes verifying the existence of a limit more intricate.

The function \( f(x, y) = 2x^2 + 3y^2 \) must satisfy the limit condition from any path converging to \((0, 0)\). A common way to handle this is by bounding the function by expressions of \( x \) and \( y \). We derived that \(|2x^2 + 3y^2| \leq 5(x^2 + y^2)\), helping to connect the \( \epsilon \) and \( \delta \) directly.

Conclusively, analyzing how close \( (x, y) \) need to be to \( (0, 0) \) for \( |f(x, y) - 0| < \epsilon \) leads us to choose a suitable \( \delta \). This ensures the function's limit evaluation is both precise and consistent regardless of the path taken.
Polar Coordinates
Polar coordinates offer a powerful tool to simplify complex limit problems, especially in multivariable calculus. By converting Cartesian coordinates \((x, y)\) into polar form using the transformations \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \), the analysis often becomes more manageable. Here, \( r \) represents the radial distance from the origin, and \( \theta \) the angle from the positive x-axis.

In this problem, noticing that \( x^2 + y^2 = r^2 \) allows us to rewrite the function \( f(x, y) = 2x^2 + 3y^2 \) as a function of \( r \). The expression \( 5(x^2 + y^2) = 5r^2 \) clearly shows the direct dependence of the function value on the radial distance \( r \) from the origin.

This simplification highlights the direct relationship inherent in the \( \epsilon \)-\( \delta \) proof, simplifying the process of finding an appropriate \( \delta \) given \( \epsilon \). Using polar coordinates is a staple technique when exploring limits in a two-dimensional context, offering clearer insights and often reducing algebraic complexity.

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Most popular questions from this chapter

A chemical diffuses in a container that occupies the interval \(0 \leq x \leq 1\). The concentration of the chemical at time \(t\) and at a point \(x\) is given by the diffusion equation: $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}} $$ (a) Suppose that the chemical is allowed to diffuse through the entire container until the concentration reaches an equilibrium value where \(c\) does not change any more with time, that is, \(\partial c / \partial t=0 .\) Suppose that chemical that touches the walls of the container is removed so that $$ c(0, t)=c(1, t)=0 . $$ The steady state concentration of chemical will be a function \(C(x)\) with $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ and \(C(0)=C(1)=0\). Show that \(C(x)=0\) satisfies this differential equation and the constraints as the points \(x=0\) and \(x=1\). (b) Now suppose that chemical is added to the container by a reaction that occurs at the wall \(x=0 .\) This reaction keeps the concentration of chemical at this wall equal to \(c(0, t)=1 . \mathrm{Un}\) der these conditions the steady state distribution of chemical will obey a differential equation: $$ 0=D \frac{d^{2} C}{d x^{2}} \text { for } x \in(0,1) $$ with \(C(0)=1\) and \(C(1)=0 .\) Show that \(C(x)=1-x\) satisfies both the differential equation and the boundary conditions at \(x=0\) and \(x=1\). (c) Notice that the steady state distributions in (a) and (b) do not depend on \(D .\) Can you explain why?

Let $$f_{a}(x, y)=a x^{2}+y^{2}$$ for \((x, y) \in \mathbf{R}\), where \(a\) is a positive constant. (a) Assume that \(a=1\) and describe the level curves of \(f_{1}\). The graph of \(f_{1}(x, y)\) intersects both the \(x-z\) and the \(y-z\) planes; show that these two curves of intersection are parabolas. (b) Assume that \(a=4\). Then $$f_{4}(x, y)=4 x^{2}+y^{2}$$ and the level curves satisfy $$4 x^{2}+y^{2}=c$$ Use a graphing calculator to sketch the level curves for \(c=\) \(0,1,2,3\), and \(4 .\) These curves are ellipses. Find the curves of intersection of \(f_{4}(x, y)\) with the \(x-z\) and the \(y-z\) planes. (c) Repeat (b) for \(a=1 / 4\). (d) Explain in words how the surfaces of \(f_{a}(x, y)\) change when \(a\) changes.

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Find the indicated partial derivatives. \(g(s, t)=\ln \left(s+t^{2}\right) ; \frac{\partial^{2} g}{\partial s^{2}}\)

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