Question

If \(\lim _{x \rightarrow x_{0}} y(x)=y_{0}\) and \(\lim _{x \rightarrow x_{0}} f(x, y(x))=L,\) we say that \(f(x, y) a p-\) proaches \(L\) as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right)\) along the curve \(y=y(x)\). (a) Prove: If \(\lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=L,\) then \(f(x, y)\) approaches \(L\) as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right)\) along any curve \(y=y(x)\) through \(\left(x_{0}, y_{0}\right)\) (b) We saw in Example 5.2 .3 that if $$ f(x, y)=\frac{x y}{x^{2}+y^{2}}, $$ then \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) does not exist. Show, however, that \(f(x, y)\) approaches a value \(L_{a}\) as \((x, y)\) approaches (0,0) along any curve \(y=y(x)\) that passes through (0,0) with slope \(a\). Find \(L_{a}\). (c) Show that the function $$ g(x, y)=\frac{x^{3} y^{4}}{\left(x^{2}+y^{6}\right)^{3}} $$ approaches 0 as \((x, y)\) approaches (0,0) along a curve as described in (b), but that \(\lim _{(x, y) \rightarrow(0,0)} f(x, y)\) does not exist.

Short answer

#Question# Prove that if the limit of a function f(x, y) exists as (x, y) approaches a point (x0, y0), then it will still approach the same limit along any curve y = y(x) through (x0, y0).

Step-by-step solution

Definition of limit

First, recall the definition of a limit: \(\lim_{(x, y) \rightarrow (x_0, y_0)} f(x, y) = L\) means that for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that $$0 < \sqrt{(x - x_0)^2 + (y - y_0)^2} < \delta \implies |f(x, y) - L| < \epsilon .$$

Using the definition of limit

Consider any curve \(y = y(x)\) passing through \((x_0, y_0)\). We need to show that along this curve, the condition for the limit stated above still holds. Let \(g(x) = f(x, y(x))\). We have $$\lim_{x \rightarrow x_0} g(x) = \lim_{x \rightarrow x_0} f(x, y(x)) = L,$$ where the last equality comes from the fact that \(\lim_{(x, y) \rightarrow (x_0, y_0)} f(x, y) = L\). Thus, for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that $$0 < |x - x_0| < \delta \implies |g(x) - L| < \epsilon .$$ Since \(g(x) = f(x, y(x))\), this means that \(f(x, y(x))\) approaches \(L\) as \((x, y(x))\) approaches \((x_0, y_0)\) along any curve \(y = y(x)\) through \((x_0, y_0)\). Thus, we have proven the statement. (b):

Finding the limit along the curve y = ax

Consider the curve \(y = ax\), where a is a constant representing the slope. The function becomes $$f(x, ax) = \frac{ax^2}{x^2 + a^2x^2}.$$ Now, let's find the limit as \((x, y)\) approaches (0, 0) along this curve: $$\lim_{x \rightarrow 0} f(x, ax) = \lim_{x \rightarrow 0} \frac{ax^2}{x^2 + a^2x^2} = \lim_{x \rightarrow 0} \frac{a}{1 + a^2} = \frac{a}{1 + a^2}.$$

Conclusion on the Limit

The limit of the function \(f(x, y) = \frac{xy}{x^2 + y^2}\) approaches \(L_a = \frac{a}{1 + a^2}\) as \((x, y)\) approaches (0, 0) along any curve \(y = ax\) that passes through (0, 0) with slope a. (c):

Find limit along the curve y = ax

Consider the curve \(y = ax\). The function becomes $$g(x, ax) = \frac{x^3 a^4 x^4}{(x^2 + a^6 x^6)^3}.$$ Now, let's find the limit as \((x, y)\) approaches (0, 0) along this curve: $$\lim_{x \rightarrow 0} g(x, ax) = \lim_{x \rightarrow 0} \frac{a^4 x^7}{(1 + a^6 x^4)^3} = 0$$

Show the limit does not exist

We will use polar coordinates to determine the limit. Let \(x = r \cos \theta\) and \(y = r \sin \theta\). Then, we can rewrite the function as $$g(r \cos \theta, r \sin \theta) = \frac{r^7 \cos^3 \theta \sin^4 \theta}{(r^2 \cos^2 \theta + r^6 \sin^6 \theta)^3}.$$ Now, let's find the limit as \((x, y) = (r \cos \theta, r \sin \theta) \rightarrow (0, 0)\): $$\lim_{r \rightarrow 0} g(r \cos \theta, r \sin \theta) = \lim_{r \rightarrow 0} \frac{r^7 \cos^3 \theta \sin^4 \theta}{(r^2 \cos^2 \theta + r^6 \sin^6 \theta)^3}.$$ This limit depends on \(\theta\), so the limit does not exist as \((x, y) \rightarrow (0, 0)\). Therefore, the function \(g(x, y) = \frac{x^3 y^4}{(x^2 + y^6)^3}\) approaches 0 as \((x, y)\) approaches (0, 0) along a curve as described in (b), but the limit does not exist as (x, y) approaches (0, 0).

Key concepts

Multivariable Limits

In calculus, multivariable limits deal with functions that depend on two or more variables. Unlike single-variable limits, where the limit is approached from just two directions (left and right), multivariable limits require the point to be approached from every possible direction in the space. This can make finding such limits more complicated.

To determine a multivariable limit without ambiguity, it's crucial that as the inputs simultaneously approach their respective target values from any path or direction, the output approaches the same limit value consistently.

• For instance, consider the function \(f(x, y) = \frac{xy}{x^2 + y^2} \). The goal is to determine \(\lim_{{(x, y) \to (0, 0)}} f(x, y) \).


• The challenge is that if the limit differs when approaching (0,0) via different paths (e.g., straight lines, parabolas), then the overall limit does not exist.

When examining multivariable functions, keep different approaches in mind. This will often lead to unexpected results. If you find a single approach that yields a different limit than all others, then the overall limit cannot be defined.

Epsilon-Delta Definition

The epsilon-delta definition is a rigorous method of defining the limit of a function at a particular point. This definition is vital in calculus to understand the core concept of limits.

When applied to multivariable functions, it means that for every positive number \(\epsilon\), no matter how small, there exists a positive number \(\delta\), such that if the distance between the point \((x, y)\) and \((x_0, y_0)\) is less than \(\delta\), then \(|f(x, y) - L|\) is less than \(\epsilon\).

• For example, if we apply this to our multivariable function \(f(x, y)\), it ensures that if we approach \((x_0, y_0)\) from any conceivable direction, the function remains within close proximity to the limit value \(L\).


• This allows for a consistent understanding and application of limits, particularly when assessing complex functions.

Understanding this definition is key to proving many limit statements in calculus, ensuring that no "path" can lead to a different result.

Polar Coordinates

Using polar coordinates is a useful strategy when dealing with limits in multivariable calculus. In polar coordinates, any point in the plane is described in terms of its radius \(r\) from the origin and its angle \(\theta\) with respect to the positive x-axis.

When calculating limits, converting a function from Cartesian to polar coordinates can often simplify the problem, especially near the origin where the functions might not be well behaved in Cartesian coordinates.

• For instance, replace \(x\) and \(y\) with \(r \, \cos \theta\) and \(r \, \sin \theta\) respectively. This can make it easier to see how the function behaves as \(r \to 0\).


• In our study of the function \(g(x, y)\), transforming to polar coordinates allowed us to demonstrate dependence on the angle \(\theta\), ultimately showing that the limit does not exist.

Using polar coordinates can help identify paths of convergence or divergence, revealing insights that are not immediately obvious in rectangular coordinates.

Curve Approaching Limit

The concept of a curve approaching a limit involves analyzing how a function behaves as its inputs approach a specific point along a particular path or curve. This is crucial when dealing with multivariable limits.

To understand limits in this context, one must consider not just straight line paths, but any potential curve that could pass through the point of interest.

• For example, approaching the origin along the curve defined by \(y = ax\) means analyzing the limit as \(x\) approaches zero for the function \(f(x, ax)\).


• If different curves yield different limit values, it suggests the limit at the point does not exist.

Curves provide a broader perspective on how a function can behave as it nears a critical point, offering a more comprehensive view of its behavior beyond linear paths.