%2F(x%5E2%2By%5E2)-limit.jpeg "(x^3+y^3)/(x^2+y^2) Limit")
Limit of (x^3+y^3)/(x^2+y^2) as (x,y) approaches (0,0)
In this article, we will explore the limit of the expression (x^3+y^3)/(x^2+y^2) as (x,y) approaches (0,0). This limit is a classic example of a multivariable limit and is often used to demonstrate the subtleties of limit calculations.
The Given Expression
The given expression is:
$\frac{x^3+y^3}{x^2+y^2}$
Approaching the Limit
As (x,y) approaches (0,0), we want to find the limit of the expression. To do this, we can start by analyzing the behavior of the numerator and denominator separately.
Numerator
The numerator is:
$x^3+y^3$
As (x,y) approaches (0,0), the numerator approaches 0. This is because both x and y are approaching 0, and the cube of a small number is also small.
Denominator
The denominator is:
$x^2+y^2$
As (x,y) approaches (0,0), the denominator also approaches 0. This is because both x and y are approaching 0, and the square of a small number is also small.
The Problem
The problem arises because both the numerator and denominator are approaching 0 as (x,y) approaches (0,0). This makes it difficult to determine the limit of the expression.
Solving the Limit
One way to solve this limit is to use the following trick:
$\frac{x^3+y^3}{x^2+y^2} = \frac{x^3+y^3}{(x^2+y^2)}\cdot\frac{x+y}{x+y}$
This may seem like a strange step, but it helps to simplify the expression. Now, we can rewrite the expression as:
$\frac{x^3+y^3}{x^2+y^2} = \frac{(x+y)(x^2-x y+y^2)}{x^2+y^2}\cdot\frac{x+y}{x+y}$
Simplifying further, we get:
$\frac{x^3+y^3}{x^2+y^2} = \frac{x^2-xy+y^2}{x+y}$
Evaluating the Limit
Now, as (x,y) approaches (0,0), we can evaluate the limit. The numerator approaches 0, and the denominator approaches 0. However, the ratio of the numerator to the denominator approaches 0/0, which is an indeterminate form.
Using L'Hopital's rule, we can differentiate the numerator and denominator separately and then evaluate the limit.
$\lim_{(x,y)\to(0,0)}\frac{x^2-xy+y^2}{x+y} = \lim_{(x,y)\to(0,0)}\frac{2x-y}{1} = 0$
Therefore, the limit of the expression (x^3+y^3)/(x^2+y^2) as (x,y) approaches (0,0) is 0.
Conclusion
In this article, we explored the limit of the expression (x^3+y^3)/(x^2+y^2) as (x,y) approaches (0,0). We used a clever trick to simplify the expression and then evaluated the limit using L'Hopital's rule. The final answer is 0, which may seem counterintuitive at first but is a true result of the calculation.
