Limit Exercise in Two Variables 2
Let's find the limit of the following function \( f(x,y) \):
$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^2y}{x^2+y^2} $$
This limit presents an indeterminate form, 0/0:
$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^2y}{x^2+y^2} = \frac{0}{0} $$
To analyze it, I'll compare the function \( f(x,y) \) with its absolute value.
Note. For any function \( f(x,y) \), it holds that $$ -| f(x,y) | \le f(x,y) \le | f(x,y) |. $$ If the limit of \( |f(x,y)| \) is zero, then by the squeeze theorem, the limit of \( f(x,y) \) is also zero: $$ \lim_{(x,y) \rightarrow (0,0)} -| f(x,y) | \le \lim_{(x,y) \rightarrow (0,0)} f(x,y) \le \lim_{(x,y) \rightarrow (0,0)} | f(x,y) | $$ $$ 0 \le \lim_{(x,y) \rightarrow (0,0)} f(x,y) \le 0. $$
Thus, we need to determine whether the limit of the absolute value is zero.
$$ \lim_{(x,y) \rightarrow (0,0)} \left| \frac{x^2y}{x^2+y^2} \right| $$
Clearly, the absolute value of the function is non-negative:
$$ 0 \le \left| \frac{x^2y}{x^2+y^2} \right| $$
We can rewrite the expression equivalently as:
$$ 0 \le \lim_{(x,y) \rightarrow (0,0)} |y| \cdot \frac{x^2}{x^2+y^2} $$
Notice that the factor \( \frac{x^2}{x^2+y^2} \) always lies between 0 and 1, since the denominator is greater than or equal to the numerator, and both are positive.
Thus, \( |y| \) is multiplied by a number between 0 and 1.
As a result, the function is bounded above by \( |y| \):
$$ 0 \le |y| \cdot \frac{x^2}{x^2+y^2} \le |y| $$
Taking the limit across the inequality:
$$ \lim_{(x,y) \rightarrow (0,0)} 0 \le \lim_{(x,y) \rightarrow (0,0)} |y| \cdot \frac{x^2}{x^2+y^2} \le \lim_{(x,y) \rightarrow (0,0)} |y| $$
The outer limits clearly tend to zero:
$$ 0 \le \lim_{(x,y) \rightarrow (0,0)} |y| \cdot \frac{x^2}{x^2+y^2} \le 0 $$
By the squeeze theorem, it follows that:
$$ \lim_{(x,y) \rightarrow (0,0)} |y| \cdot \frac{x^2}{x^2+y^2} = 0 $$
and therefore:
$$ \lim_{(x,y) \rightarrow (0,0)} \left| \frac{x^2y}{x^2+y^2} \right| = 0 $$
Consequently, the limit of the original function must also be zero:
$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^2y}{x^2+y^2} = 0 $$
Explanation. By the squeeze theorem: $$ \lim_{(x,y) \rightarrow (0,0)} - \left| \frac{x^2y}{x^2+y^2} \right| = 0 \le \lim_{(x,y) \rightarrow (0,0)} \frac{x^2y}{x^2+y^2} \le \lim_{(x,y) \rightarrow (0,0)} \left| \frac{x^2y}{x^2+y^2} \right| = 0 $$
Alternative Approach
We can also compute the limit by switching to polar coordinates.
\[ \lim_{(x,y) \to (0,0)} \frac{x^2 y}{x^2 + y^2} \]
Why use polar coordinates? Polar coordinates are particularly helpful when studying limits as \((x,y)\) approaches \((0,0)\), because they simplify the expression and make it easier to track how the function depends on the distance from the origin.
The change to polar coordinates is given by:
\[ x = \rho \cos \theta \]
\[ y = \rho \sin \theta \]
where \(\rho\) represents the distance from the origin (\( \rho = \sqrt{x^2 + y^2} \)) and \(\theta\) is the angle relative to the positive \(x\)-axis.
Substituting these into the function yields:
\[ \frac{x^2 y}{x^2 + y^2} = \frac{(\rho \cos\theta)^2 (\rho \sin\theta)}{(\rho \cos\theta)^2 + (\rho \sin\theta)^2} \]
Expanding and simplifying:
\[ \frac{x^2 y}{x^2 + y^2} = \frac{\rho^2 \cos^2\theta \cdot \rho \sin\theta}{\rho^2 \cos^2\theta + \rho^2 \sin^2\theta} \]
\[ = \frac{\rho^3 \cos^2\theta \sin\theta}{\rho^2 (\cos^2\theta + \sin^2\theta)} \]
By the Pythagorean identity, \( \cos^2\theta + \sin^2\theta = 1 \), so this simplifies to:
\[ \frac{x^2 y}{x^2 + y^2} = \frac{\rho^3 \cos^2\theta \sin\theta}{\rho^2} \]
Thus, the function becomes:
\[ \frac{x^2 y}{x^2 + y^2} = \rho \cos^2\theta \sin\theta \]
We now take the limit:
\[ \lim_{\rho \to 0} \rho \cos^2\theta \sin\theta \]
Since \(\theta\) is held constant as \(\rho\) tends to zero, \( \cos^2\theta \sin\theta \) acts as a constant.
Thus, the product tends to zero as \(\rho\) goes to zero:
\[ \cos^2\theta \sin\theta \cdot \lim_{\rho \to 0} \rho = 0 \]
For full rigor, we can apply the squeeze theorem, observing that \( \cos^2\theta \sin\theta \) is bounded between \(-1\) and \(1\).
\[ -\rho \le \rho \cos^2\theta \sin\theta \le \rho \]
Taking limits as \( \rho \to 0 \):
\[ \lim_{\rho \to 0} -\rho \le \lim_{\rho \to 0} \rho \cos^2\theta \sin\theta \le \lim_{\rho \to 0} \rho \]
\[ -\lim_{\rho \to 0} \rho \le \lim_{\rho \to 0} \rho \cos^2\theta \sin\theta \le \lim_{\rho \to 0} \rho \]
Since \( \lim_{\rho \to 0} \rho = 0 \), it follows that:
\[ \lim_{\rho \to 0} \rho \cos^2\theta \sin\theta = 0 \]
Thus, the alternative method confirms the result more cleanly by expressing the dependence on the distance from the origin via the variable \(\rho\).
In conclusion, the limit of the two-variable function is:
\[ \lim_{(x,y)\to(0,0)} \frac{x^2 y}{x^2 + y^2} = 0 \]
And that wraps it up.
