An lower bound for 1+x2 is definitely 1.trap101 said:Use the delta-epsilon definition to prove f(x,y) = y/(1+x2) is continuous at (0,0)
Attempts:
So I'm doing some work and my main issue is finding a bound for the denominator of 1+x2:
So work wise I have something looking like:
[itex]\delta[/itex]/(|1| + |x2| ). How could I found a good bound?
The Delta-Epsilon bound is a mathematical concept used in calculus and analysis to prove the limit of a function. It is also known as the Epsilon-Delta definition of the limit.
The Delta-Epsilon bound states that for a given function f(x), if there exists a positive number δ (delta) such that for all values of x within a distance of δ from a point a, the difference between f(x) and the limit L is less than a positive number ε (epsilon), then the limit of f(x) as x approaches a is equal to L.
The Delta-Epsilon bound is important because it provides a rigorous and precise definition of the concept of a limit, which is a fundamental concept in calculus. It allows for the calculation of limits for more complex functions and provides a foundation for many other concepts in calculus, such as continuity and differentiability.
The Delta-Epsilon bound is used in real-world applications, such as engineering and physics, to analyze and model real-world phenomena. It allows for the determination of critical values and helps to understand the behavior of a system as it approaches a certain point.
While the Delta-Epsilon bound is a powerful tool in calculus, it does have its limitations. It can only be used to prove the existence of a limit, but it does not provide a method for calculating the limit. Additionally, it only applies to functions with real numbers as inputs and outputs and cannot be used for functions with complex numbers.