Evaluating the Limit: ( lim_{(x, y) to (0, 0)} frac{x^3 - xy^2}{x^2 y^2} )

In this article, we will explore the evaluation of a limit problem, specifically ( lim_{(x, y) to (0, 0)} frac{x^3 - xy^2}{x^2 y^2} ). We will approach this problem using both direct methods and polar coordinates, emphasizing the use of the Squeeze Theorem.

Solution Using Direct Methods and Squeeze Theorem

First, let's consider the given limit:

( lim_{(x, y) to (0, 0)} frac{x^3 - xy^2}{x^2 y^2} )

One can approach this problem by using the triangle inequality and the Squeeze Theorem. For all (x, y eq 0, 0), we have:

( left| frac{x^3 - xy^2}{x^2 y^2} right| leq left| frac{x^3}{x^2 y^2} right| left| frac{-xy^2}{x^2 y^2} right| )

Using the triangle inequality, we can bound each term:

( left| frac{x^3}{x^2 y^2} right| leq |x| )

( left| frac{-xy^2}{x^2 y^2} right| leq |x| )

Thus, we have:

( left| frac{x^3 - xy^2}{x^2 y^2} right| leq |x| |x| 2|x| )

Since

( lim_{(x, y) to (0, 0)} 2|x| 0 )

We conclude by the Squeeze Theorem that:

( lim_{(x, y) to (0, 0)} frac{x^3 - xy^2}{x^2 y^2} 0 )

Solution Using Polar Coordinates

Another method to solve this problem is by switching to polar coordinates. Let

( x r cos(theta) )

( y r sin(theta) )

Substituting into the limit, we get:

( lim_{r to 0} frac{r^3 cos^3(theta) - r cos(theta) sin^2(theta)}{r^2} lim_{r to 0} left( r cos^3(theta) - cos(theta) sin^2(theta) right) )

Observing that for every ( theta in mathbb{R} ) the inequality holds:

( 0 leq cos^3(theta) - cos(theta) sin^2(theta) leq 2 )

This implies:

( -2r leq r cos^3(theta) - cos(theta) sin^2(theta) leq 2r )

We can now apply the Squeeze Theorem to conclude that:

( lim_{r to 0} left( r cos^3(theta) - cos(theta) sin^2(theta) right) 0 )

Thus, the limit is 0.

Alternative Method Using Direct Substitution and Simplification

Alternatively, we can rewrite the numerator as:

( x^3 - xy^2 x(x^2 - y^2) )

Thus:

( frac{x^3 - xy^2}{x^2 y^2} frac{x(x^2 - y^2)}{x^2 y^2} )

We can further simplify using the fact that:

( x^2 - y^2 (x y)(x - y) )

So:

( frac{x(x^2 - y^2)}{x^2 y^2} x cdot frac{(x y)(x - y)}{x^2 y^2} )

Since:

( left| frac{(x y)(x - y)}{x^2 y^2} right| leq 2 )

We conclude that:

( -2x leq x cdot frac{(x y)(x - y)}{x^2 y^2} leq 2x )

And:

( lim_{(x, y) to (0, 0)} 2x 0 )

Hence, by the Squeeze Theorem:

( lim_{(x, y) to (0, 0)} frac{x^3 - xy^2}{x^2 y^2} 0 )

This establishes the limit.