Question

Find (if possible) the ordinary, the double, and the iterated limits of \(f\) at (0,0) assuming that \(f(x, y)\) is given by one of the expressions below, and \(f\) is defined at those points of \(E^{2}\) where the expression has sense. (i) \(\frac{x^{2}}{x^{2}+y^{2}} ;\) (ii) \(\frac{y \sin x y}{x^{2}+y^{2}} ;\) (iii) \(\frac{x+2 y}{x-y}\); (iv) \(\frac{x^{3} y}{x^{6}+y^{2}}\) (v) \(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\) (vi) \(\frac{x^{5}+y^{4}}{\left(x^{2}+y^{2}\right)^{2}} ;\) (vii) \(\frac{y+x \cdot 2^{-y^{2}}}{4+x^{2}}\) (viii) \(\frac{\sin x y}{\sin x \cdot \sin y}\).

Short answer

Ordinary limits exist only for (ii) and (vii), both are 0.

Step-by-step solution

Evaluate the Ordinary Limit for Expression (i)

We begin with the expression \( f(x, y) = \frac{x^{2}}{x^{2} + y^{2}} \). To find the limit as \((x, y) \to (0,0)\), we choose the path \(y = 0\). Substituting, we get \( f(x, 0) = \frac{x^2}{x^2} = 1 \). Now choose the path \(x = 0\), then \( f(0, y) = \frac{0}{y^2} = 0 \). Since the results differ, the ordinary limit does not exist.

Evaluate the Ordinary Limit for Expression (ii)

For \( f(x, y) = \frac{y \sin(xy)}{x^{2} + y^{2}} \), we check the limit by the path \(y = 0\): substituting gives \(f(x, 0) = 0\). For the path \(x = 0\), we get \(f(0, y) = 0\). Let’s check by the path \(y = x\): we get \(f(x, x) = \frac{x \sin(x^2)}{2x^2} = \frac{\sin(x^2)}{2x}\), which goes to zero as \(x \to 0\). The ordinary limit is 0.

Evaluate the Ordinary Limit for Expression (iii)

Given \( f(x, y) = \frac{x + 2y}{x - y} \), try the path \(y = x \): the expression becomes undefined as it simplifies to division by zero. Since the limit depends on the path chosen and some paths lead to undefined expressions, the ordinary limit does not exist.

Evaluate the Ordinary Limit for Expression (iv)

For \( f(x, y) = \frac{x^{3} y}{x^{6} + y^{2}} \), consider the path \(y = 0\): \(f(x, 0) = 0\). For \(x = 0\), \(f(0, y) = 0\). By the path \(y = x^{3}\), it simplifies to \(\frac{x^3 x^3}{x^6 + x^6} = \frac{x^6}{2x^6} = \frac{1}{2}\). The ordinary limit does not exist.

Evaluate the Ordinary Limit for Expression (v)

For \( f(x, y) = \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \), try both paths \(y = 0\) and \(x = 0\), the limits in both cases simplify to \(1\) and \(-1\) respectively. Thus, the ordinary limit does not exist.

Evaluate the Ordinary Limit for Expression (vi)

For \( f(x, y) = \frac{x^5 + y^4}{(x^2 + y^2)^2} \), using path \(y = 0\), we get \(f(x,0) = \frac{x^5}{x^4} = x\), approaching zero as \(x \to 0\). For \(x = 0\), we have \(f(0,y) = \frac{y^4}{y^4} = 1\). The limit by this path doesn't match others; thus, the ordinary limit does not exist.

Evaluate the Ordinary Limit for Expression (vii)

For \( f(x, y) = \frac{y + x \cdot 2^{-y^{2}}}{4 + x^{2}} \), simplify using \(y = 0\): \(f(x,0) = \frac{x}{4+x^2}\), approaches zero. For \(x = 0\), \(f(0,y)= \frac{y}{4}\), also approaches zero. Ordinary limit exists and is 0.

Evaluate the Ordinary Limit for Expression (viii)

For \( f(x, y) = \frac{\sin(xy)}{\sin x \cdot \sin y} \), substituting \((0, 0)\), the function is undefined. Evaluating along path \(y = x\) becomes \(\frac{\sin(x^2)}{\sin^2 x}\), which does not simplify to a single value as \(x \to 0\). Ordinary limit does not exist.

Key concepts

Ordinary Limit

The ordinary limit, sometimes simply referred to as 'the limit,' involves evaluating the behavior of a two-variable function as one approaches a particular point, in this case (0,0), from any direction in its domain. To establish the existence of an ordinary limit, the function must yield the same result irrespective of the path taken. If different paths yield different outcomes, the ordinary limit does not exist.

• Pick various paths like setting one of the variables to zero or equating them.
• Calculate the resulting limit for each chosen path.

For example, function (i) \( \frac{x^2}{x^2 + y^2} \) gave path-dependent results: it reached 1 when \( y = 0 \) but 0 when \( x = 0 \). This indicates an inconsistent approach, thus the ordinary limit does not exist.

Double Limit

A double limit examines the simultaneous approach of both variables to a point. This type is slightly more challenging since you have to maintain control over both variables' behavior.
In practice, one typically analyzes both the horizontal and the vertical limits, like \( x \to 0 \) and \( y \to 0 \). If each limit holds and converges to the same value, the double limit exists. If not, or if any path leads to a different value, it does not exist. In our exercises, explicit double limit calculations weren't presented, but they require an extension of path analysis to consider the strict conditions under which both variables converge.

• Evaluate horizontal and vertical limit pathways.
• Ensure consistency across all examined paths.

The challenge with double limits is ensuring every conceivable path aligns.

Iterated Limit

Iterated limits involve narrowing down the behavior of a function by successively taking limits in specific sequences of its variables. Unlike a double limit, where both variables shift concurrently, iterated limits involve a stepwise sequence: first holding one constant and varying the other, then switching.
The sequence in which these limits are evaluated influences the result.

• First fix one value and find its limit as the other approaches zero.
• Then switch the roles, and repeat.

A distinct result or dependency on the order of limits implies no unified limit exists. In the context of provided exercises, iterated limits were tested by fixing either \( x = 0 \) or \( y = 0 \) first, subsequently examining the alternate variable's behavior, such as demonstrated with \( \frac{y \sin(xy)}{x^2 + y^2} \) and observing its behavior along specific paths. This can clarify if solutions exist uniquely or depend on variable sequences.