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Chapter 11: Problem 7

Find the limit, if it exists, or show that the limit does not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{y^{2} \sin ^{2} x}{x^{4}+y^{4}}$$

Short Answer

Expert verified
The limit does not exist as different paths yield different results.

Step by step solution

01

Understanding the Limit

We're given a two-variable function and need to find if the limit exists as \((x, y)\) approaches \((0, 0)\). The function is \(\frac{y^2 \sin^2 x}{x^4 + y^4}\). We will analyze this by approaching along various paths to check if it yields the same result. If we get different results, the limit does not exist.
02

Approaching Along the y-axis

First, consider the path \(x = 0\). Substituting \(x = 0\) in the expression, we get: \[f(0, y) = \frac{y^2 \sin^2 0}{0^4 + y^4} = 0\] because \(\sin 0 = 0\). Thus, the limit along the \(y\)-axis is 0.
03

Approaching Along the x-axis

Consider the path \(y = 0\). Substituting \(y = 0\) in the expression, we obtain: \[f(x, 0) = \frac{0^2 \sin^2 x}{x^4 + 0^4} = 0\] Thus, the limit along the \(x\)-axis is also 0.
04

Approaching Along a Line y = mx

Now let's consider the path \(y = mx\). Substituting \(y = mx\) into the expression gives: \[f(x, mx) = \frac{(mx)^2 \sin^2 x}{x^4 + (mx)^4} = \frac{m^2x^2 \sin^2 x}{x^4(1 + m^4)} = \frac{m^2 \sin^2 x}{x^2(1 + m^4)}\] As \(x \to 0\), since \(\sin^2 x \approx x^2\), the expression simplifies to \(\frac{m^2 x^2}{x^2(1 + m^4)} = \frac{m^2}{1 + m^4}\). This shows that by varying \(m\), different limits can be obtained, meaning the overall limit doesn't exist.

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Key Concepts

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

Two-variable functions
Working with functions of two variables can be an exciting venture into the world of multivariable calculus. A two-variable function is essentially a rule that assigns a specific output value to every possible ordered pair \((x, y)\) of input values. In our case, we are dealing with the function \(f(x, y) = \frac{y^2 \sin^2 x}{x^4 + y^4}\). This function's output changes based on the values of both \(x\) and \(y\).
A major aspect of understanding such functions is visualizing them as surfaces in three-dimensional space. The inputs \(x\) and \(y\) live on a plane, and the output \(f(x, y)\) depicts the height of the surface above that point in the plane.
Dealing with these functions can become tricky when considering limits, as the approach to a given point can vary in infinitely many ways.
Path analysis in limits
Evaluating limits of functions with two variables is a bit different compared to single-variable limits. Path analysis is a method used to determine if a limit exists at a particular point. This involves checking what happens to the function as the inputs approach this point along different paths.
  • If the limit value is the same across all paths, it suggests (but does not confirm) that the limit might exist.
  • If different paths yield different limit values, the limit does not exist.

In our exercise, the function \(\frac{y^2 \sin^2 x}{x^4 + y^4}\) was evaluated along the axes \((x=0, y=0)\) and along lines \(y=mx\). These paths showed that variable approaches yield variable results, such as different limit values depending on \(m\). This indicates the non-existence of a single limit at the point \((0, 0)\).
Limit existence criteria
To determine if a limit exists for a multivariable function, we must consider the path-independent value. This means that the limit should not change regardless of how the point \((x, y)\) is approached.
For a multivariable limit to exist, the following criteria should be met:
  • The outcome should be the same across all paths approaching the target.
  • The function should not produce contradictory results as it approaches the desired point.

If any of these criteria are not met, like in our exercise, the limit does not exist. Here, the limit was found to vary when evaluated using different paths, such as along different lines \(y = mx\), thus confirming that the limit does not exist at \((0, 0)\).
Understanding this helps to solidify one鈥檚 grasp of multivariable calculus and the complexities involved when handling two-variable functions.

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