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Find the limits in Exercises \(1-12.\) $$\lim _{(x, y) \rightarrow(0, \ln 2)} e^{x-y}$$

Short Answer

Expert verified
The limit is \(\frac{1}{2}\).

Step by step solution

01

Understand the Limit Expression

We need to find the limit of the function \(f(x,y) = e^{x-y}\) as \((x,y)\) approaches \((0, \, \ln 2)\). This means we want to know the value that \(f(x, y)\) gets closer to as \(x\) gets closer to 0 and \(y\) gets closer to \(\ln 2\).
02

Substitute the Approaching Values

In this particular problem, we substitute \((x, y) = (0, \ln 2)\) into the expression \(e^{x-y}\). This gives us \(e^{0 - \ln 2}\).
03

Simplify the Exponent

Simplify the exponent \(0 - \ln 2\). It simplifies to \(-\ln 2\). This means we need to evaluate \(e^{-\ln 2}\).
04

Use Properties of Exponents and Logarithms

Use the property \(e^{-\ln a} = \frac{1}{a}\) for any positive number \(a\). Applying this property, we have \(e^{-\ln 2} = \frac{1}{2}\).
05

Write Down the Limit Result

Based on the simplification and properties, the limit when \((x, y)\) approaches \((0, \ln 2)\) of the function \(e^{x-y}\) is \(\frac{1}{2}\).

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Key Concepts

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

Limit of a function
A limit of a function is a fundamental concept in calculus. It describes the behavior of a function as inputs get closer to a certain point. Imagine a function like a road, and the limit is the destination you're trying to reach. As you approach a specific point, you want to know where the function is heading. Limits help figure this out even if the function doesn’t exactly reach that point.

For the given exercise, we're dealing with a multivariable limit. This means we're examining how a function behaves when two variables move towards specific values. In our case, we are interested in what happens to the function \(f(x, y) = e^{x-y}\) as \((x, y)\) approaches \((0, \ln 2)\). The limit we're finding shows the value this function approaches with these inputs, even if \(x\) and \(y\) aren't exactly 0 and \(\ln 2\), just getting very close.

Understanding limits is crucial because they are basic blocks of calculus. They form the foundation for concepts like continuity, derivatives, and integrals.
Exponential function
An exponential function is one where the variable appears in the exponent. A common form is \(e^x\), where \(e\) is a special constant approximately equal to 2.71828. Known as Euler's number, it's the backbone of exponential growth and decay processes.

In our problem, the function \(e^{x-y}\) is an example of an exponential setup. Here, the exponent is a difference of variables: \(x - y\). As these variables change, they affect the value of \(e^{x-y}\) significantly. Exponential functions are powerful because they can model situations where growth accelerates—the larger the value, the faster it grows or decays.
  • They're incredibly useful in a range of fields: natural sciences, economics, statistics, and more.
  • Often seen in scenarios involving compound interest, population growth, or radioactive decay.

Grasping how exponential functions respond to shifts in variables is key. It can demystify scenarios where they are applied and predict how they evolve.
Properties of logarithms
Logarithms are the inverse operations of exponentiation. \(\ln x\), often read as "the natural log of \(x\)," is a logarithm with the base \(e\). Understanding its properties can simplify complex problems, like the one in our exercise.

Several key properties of logarithms make them useful:
  • Product Rule: \(\ln(ab) = \ln a + \ln b\)
  • Quotient Rule: \(\ln\left(\frac{a}{b}\right) = \ln a - \ln b\)
  • Power Rule: \(\ln(a^b) = b \ln a\)
  • Identity: \(e^{\ln x} = x\) for \(x > 0\)

In our problem, we used the property that \(e^{-\ln a} = \frac{1}{a}\), applying it to find \(e^{-\ln 2} = \frac{1}{2}\). Logarithmic properties enable these kinds of simplifications by turning exponentiation problems into multiplication and division, which are more straightforward to handle.

Understanding these properties enhances mathematical flexibility, allowing scholars to dissect and solve a variety of problems not just in math but in real-world applications as well.

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Most popular questions from this chapter

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z, w)=x^{2}+y^{2}+z^{2}+w^{2}\) subject to the constraints \(\quad 2 x-y+z-w-1=0 \quad\) and \(\quad x+y-z+\) \(w-1=0.\)

Use a CAS to plot the implicitly defined level surfaces in Exercises \(73-76 .\) $$ \sin \left(\frac{x}{2}\right)-(\cos y) \sqrt{x^{2}+z^{2}}=2 $$

Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes \(y+2 z=12\) and \(x+y=6 .\)

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Suppose that the equation \(g(x, y, z)=0\) determines \(z\) as a differentiable function of the independent variables \(x\) and \(y\) and that \(g_{z} \neq 0 .\) Show that \begin{equation}\left(\frac{\partial z}{\partial y}\right)_{x}=-\frac{\partial g / \partial y}{\partial g / \partial z}.\end{equation}

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