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Find the indicated limit or state that it does not exist. \(\lim _{(x, y) \rightarrow(0,0)} \frac{\tan \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}\)

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Understand the Problem

We are tasked with finding the limit of \( \frac{\tan(x^2 + y^2)}{x^2 + y^2} \) as \( (x, y) \to (0, 0) \). This means we need to evaluate the behavior of this expression as both \( x \) and \( y \) approach 0 simultaneously.
02

Use a Substitution

To simplify, let \( r^2 = x^2 + y^2 \). As \( (x, y) \to (0, 0) \), \( r \to 0 \). This substitution transforms our limit into \( \lim_{r \to 0} \frac{\tan(r^2)}{r^2} \).
03

Apply Limit Properties

We use the limit property that \( \lim_{u \to 0} \frac{\tan(u)}{u} = 1 \). Here, set \( u = r^2 \). Then, as \( r \to 0 \), \( u \to 0 \) as well, and thus \( \lim_{r \to 0} \frac{\tan(r^2)}{r^2} = 1 \).
04

Conclude the Solution

Given that \( \frac{\tan(u)}{u} \to 1 \) as \( u \to 0 \), by substitution we found that the original limit \( \lim_{(x, y) \to (0,0)} \frac{\tan(x^2 + y^2)}{x^2 + y^2} \) is 1.

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Key Concepts

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

Limit Evaluation Techniques
Evaluating limits is a critical skill in calculus, especially when assessing behaviors near certain points. This exercise involves a multivariable function, where both variables approach zero. One effective method to simplify such problems is substitution.
Substitution reduces a complex expression to a simpler form. In this case, by letting \( r^2 = x^2 + y^2 \), the multivariable limit transforms into a single-variable problem. This simplifies our calculation significantly.
A common technique with trigonometric functions like tangent is to apply known limit properties. For instance, the property \( \lim_{u \to 0} \frac{\tan(u)}{u} = 1 \) directly aids in simplifying and evaluating these expressions. Such known limits are powerful tools that help solve problems involving indeterminate forms.
- Simplifying complex expressions with substitution helps focus on the core of the problem.- Utilizing known limits allows us to draw conclusions quickly and accurately.
Multi-Variable Calculus
In multivariable calculus, limits can be a bit more challenging due to multiple dimensions. Unlike single-variable limits that only consider changes along a line, multivariable limits consider changes in a plane or space.
Points do not simply approach zero; they do so in every direction, providing a more comprehensive check of the function's behavior. A critical newbie mistake is assuming the limit based on only one path. However, the limit has to be consistent regardless of the path taken towards the point.
Using polar or another form of substitution, particularly when the problem exhibits radial symmetry, helps simplify the evaluation as it reduces the dimensions. This transforms the problem into a more familiar one-dimensional space.
- Multivariable limits offer a broad view that requires consideration of all possible paths. - Substitutions like polar change simplify the complex 2D or 3D problem to a simpler form.
Trigonometric Limits
Trigonometric limits often involve functions which can become indeterminate. These typically require clever manipulation or the application of known trigonometric limit properties.
The limit problem we tackled uses the property \( \lim_{u \to 0} \frac{\tan(u)}{u} = 1 \). This property helps resolve the indeterminate form by comparing the rate of change in the tangent function to a linear function near zero. Deploying these properties often transforms complex limits into simpler, more standard problems.
It is crucial to recognize when traditional approach fails and a specific trigonometric identity or limit property simplifies the task. These properties are not just memorization; they are strategic tools for limit evaluation.
- Trigonometric limits can appear complex due to their indeterminate form.- Employing key limit properties simplifies these problems into manageable solutions.

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Most popular questions from this chapter

Call a function \(f(x, y)\) homogeneous of degree 1 if \(f(t x, t y)=t f(x, y)\) for all \(t>0 .\) For example, \(f(x, y)=\) \(x+y e^{y / x}\) satisfies this criterion. Prove Euler's Theorem that such a function satisfies $$ f(x, y)=x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} $$

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