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.