Question

Use the method of your choice to ate the following limits. $$\lim _{(x, y) \rightarrow(1,0)} \frac{y \ln y}{x}$$

Short answer

Answer: The value of the limit is 0.

Step-by-step solution

Analyze the function

First, let's look at the given limit: $$\lim _{(x, y) \rightarrow(1,0)} \frac{y \ln y}{x}$$ This is a limit of two variables (x, y). However, we can observe that the function can be separated into y and ln(y) terms. This indicates that we can use substitution to simplify the function.

Introduce the substitution

Let's consider the substitution: $$u =\ln y$$ Now, we can rewrite the function in terms of u, and find the corresponding limit as (x, y) approaches (1, 0). From the substitution, we can express y as: $$y = e^u$$ Now, substitute this into the given limit: $$\lim _{(x, y) \rightarrow(1,0)} \frac{y \ln y}{x} = \lim _{(x, u) \rightarrow(1, -\infty)} \frac{e^u u}{x}$$

Analyze the new limit

Now we analyze the new limit: $$\lim _{(x, u) \rightarrow(1, -\infty)} \frac{e^u u}{x}$$ The variable x approaches 1, and the variable u approaches negative infinity. We can now apply these limits individually and simplify the function.

Apply the limit to x and u separately

Applying the limit to x and u separately: $$\lim _{x \rightarrow 1} \frac{1}{x} \cdot \lim _{u \rightarrow -\infty} ue^u$$ Now we evaluate the limits: $$\lim _{x \rightarrow 1} \frac{1}{x} = 1$$ For the second limit, we can use L'Hopital's Rule since it is in the indeterminate form \(-\infty \cdot 0\): $$\lim _{u \rightarrow -\infty} ue^u = \lim _{u \rightarrow -\infty} \frac{u}{e^{-u}}$$ Using L'Hopital's Rule, differentiate both the numerator and denominator: $$\lim _{u \rightarrow -\infty} \frac{1}{-e^{-u}} = 0$$

Compute the final limit

Now, we multiply the evaluated limits: $$1 \cdot 0 = 0$$ So, the final limit is: $$\lim _{(x, y) \rightarrow(1,0)} \frac{y \ln y}{x} = 0$$

Key concepts

Multivariable Calculus

Multivariable calculus is an extension of single-variable calculus to functions of two or more variables. It focuses on understanding and evaluating functions that depend on several inputs. In this exercise, we dealt with a function of two variables, \(x\) and \(y\), and explored their behavior as they approach specific values. This is referred to as a limit in multivariable calculus.When approaching multivariable limits, you might often notice they aren't as straightforward as single-variable ones. You have to consider the function's behavior from all possible directions as it approaches the target point. This can lead to complex scenarios, but also incentivizes methods such as substitutions to streamline the process.Knowing how to cleverly transform a function, evaluate its limits using simplification techniques, and deduce the result is key to mastering multivariable calculus. With practice, you can predict and handle intricate limit problems by identifying useful patterns and methods.

L'Hopital's Rule

L'Hopital's Rule is a fundamental tool in calculus for evaluating limits that fall into indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). It states that if the limit result is indeterminate after direct substitution, it can be resolved by differentiating the numerator and the denominator separately until a determinate form emerges.In the given exercise, after substitution and transforming the variables, we utilized L'Hopital's Rule to evaluate the limit \(\lim _{u \rightarrow -\infty} \frac{u}{e^{-u}}\). The form of the limit was initially \(-\infty \cdot 0\), which required converting it to a fraction to apply L'Hopital's Rule effectively.Using L'Hopital's Rule involves:

• Converting an undefined limit into a fraction with the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
• Taking the derivatives of the numerator and denominator.
• Re-evaluating the limit after differentiation.

This technique simplifies complex limits by breaking them down into manageable steps, guiding us to find their actual value.

Substitution Method

The substitution method involves replacing complex expressions within a limit with simpler, equivalent expressions that are easier to work with. This technique was crucial in solving the exercise, transforming a multivariable limit problem into one involving single variables.For the provided limit \(\lim _{(x, y) \rightarrow(1,0)} \frac{y \ln y}{x}\), we introduced the substitution \(u = \ln y\). By expressing \(y\) in terms of \(u\), we simplified the original expression:\[y = e^u\]With the substitution, the variables transformed, enabling us to analyze the limit from a clearer perspective. The substitution reduced the multivariable problem to analyzing \(\frac{e^u u}{x}\) as \(x\rightarrow 1\) and \(u \rightarrow -\infty\), where each variable could be treated independently.This approach showcases the power of the substitution method:

• Transforms complex multivariable limits into simpler expressions.
• Makes it easier to apply calculus rules effectively.
• Allows for individual analysis of transformed variables.

Substitution is a versatile technique in calculus, helping solve challenging problems by simplifying the landscape.

Indeterminate Forms

Indeterminate forms arise when evaluating limits that yield uncertain or undefined results. Common forms include \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(\infty - \infty\), and \(0 \cdot \infty\). Recognizing these forms is essential as they indicate that further analysis or algebraic manipulation is necessary to determine the limit.In the context of our exercise, after substitution, the expression \(ue^u\) resulted in a form \(-\infty \cdot 0\), one of these indeterminate cases. This signifies that direct computation won't yield a simple solution.Handling indeterminate forms typically involves techniques like:

• L'Hopital's Rule: Differentiating the numerator and denominator.
• Algebraic Manipulation: Re-arranging terms to create determinate forms.
• Substitution: Introducing new variables to simplify the expression.

By identifying a limit's indeterminate form, you can apply suitable calculus strategies to reveal and solve the problem, as demonstrated in the given exercise. Such forms are a common hurdle in calculus but learning to navigate them is a valuable skill.