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Show that the limits do not exist. $$\lim _{(x, y) \rightarrow(1,1)} \frac{\tan y-y \tan x}{y-x}$$

Short Answer

Expert verified
The limit does not exist since varied paths yield inconsistent results.

Step by step solution

01

Understand the Problem

We need to find the limit of the expression \( \frac{\tan y - y \tan x}{y - x} \) as \((x, y)\) approaches \((1, 1)\). If the limit is ... different paths, the limit does not exist.
02

Approach Along Path 1

First, check the limit as \( y = x \). Substitute \( y = x \) into the expression: \[\frac{\tan x - x \tan x}{x - x} = \frac{0}{0},\]which is an indeterminate form. We need to explore another path to conclude if it might not exist.
03

Approach Along Path 2

Try an alternative path, \( y = x^2 \). Substitute \( y = x^2 \) into the expression:\[\frac{\tan x^2 - x^2 \tan x}{x^2 - x}.\]Examine this further using L'Hôpital's Rule or by simplifying, if possible, to resolve the indeterminate form.
04

Apply L'Hôpital's Rule on Path 2

Since \( \frac{\tan x^2 - x^2 \tan x}{x^2 - x} \) is still indeterminate, apply L'Hôpital's Rule:Differentiate the numerator and the denominator with respect to \(x\):- Numerator: \( d/dx[\tan(x^2) - x^2 \tan x] \)- Denominator: \( d/dx[x^2 - x] = 2x - 1 \)Continue simplifying until you resolve this or check for inconsistencies as \(x \rightarrow 1\).
05

Check Different Paths

To conclude the limit doesn't exist, use another path. Try \( y = 1 \) and \( x \rightarrow 1 \):\[\frac{\tan 1 - 1 \cdot \tan x}{1 - x}.\]See if it provides a different result or leads to a contradiction.
06

Final Conclusion

Upon taking these different paths, if the limit gives varying results or fails to converge to a single value consistently, you can conclude that the limit as \((x, y) \to (1, 1)\) does not exist.

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Key Concepts

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

L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool for evaluating limits that initially yield indeterminate forms, like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It provides a method to simplify these forms by differentiating the numerator and the denominator separately.

Here's how you can apply it:
  • First, confirm the limit is an indeterminate form.
  • Differentiation: Differentiate the top and bottom separately.
  • Re-evaluate the limit: Check if it can be easily computed after differentiation.
In the exercise, L'Hôpital's Rule is used on the expression \( \frac{\tan x^2 - x^2 \tan x}{x^2 - x} \). Both the numerator \( \tan(x^2) - x^2 \tan x \) and denominator \( x^2 - x \) are differentiated with respect to \( x \). This has the potential to simplify the fraction, making it easier to solve as \( x \) approaches 1.

However, remember, applying L'Hôpital's Rule is only helpful if subsequent evaluations do not result in another indeterminate form or if they help to establish path dependence later on.
Indeterminate forms
Indeterminate forms occur when the direct substitution of values into an expression results in undefined or ambiguous values. Common examples include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), among others.

Why are they problematic? They do not offer clear information about the behavior of a function as it nears a limit. In the problem, the specific indeterminate form \( \frac{0}{0} \) arises when substituting various paths into the limit expression. This signals the need for alternative strategies, such as solution by L'Hôpital's Rule or path analysis, to try and resolve the form.

Indeterminate forms invite us to try different approaches until a clear resolution can be determined. They challenge us to investigate the expressions deeper, posing a puzzle that often requires more subtle analysis or manipulation to solve.
Path dependence in limits
Path dependence in limits refers to situations where the limit value can vary depending on the path taken to approach a particular point in the domain of a function. Limits that depend on the path indicate that the limit at that point does not exist.

In this exercise, the limit \( \lim _{(x, y) \rightarrow(1,1)} \frac{\tan y-y \tan x}{y-x} \) is investigated along different paths like \( y = x \) and \( y = x^2 \). Each path provided indeterminate outcomes, encouraging further exploration. Eventually, differing results from these paths can demonstrate that a consistent limit is unreachable.

The concept of path dependence is crucial when dealing with multivariable limits. It highlights how the approach or trajectory matters when evaluating limits in higher-dimensional spaces. If different approaches lead to distinct values or no conclusive result, it supports the conclusion that a limit does not exist in that context.

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Most popular questions from this chapter

Consider the function \(f(x, y)=x^{2}+y^{2}+2 x y-x-y+1\) over the square \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\) a. Show that \(f\) has an absolute minimum along the line segment \(2 x+2 y=1\) in this square. What is the absolute minimum value? b. Find the absolute maximum value of \(f\) over the square.

Sum of products \(\operatorname{Let} a_{1}, a_{2}, \ldots, a_{n}\) be \(n\) positive numbers. Find the maximum of \(\Sigma_{i=1}^{n} a_{i} x_{i}\) subject to the constraint \(\Sigma_{i=1}^{n} x_{i}^{2}=1\)

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Show that each function in Exercises \(83-90\) satisfies a Laplace equation. $$f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}$$

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