# Limit of a Nonnegative Continuous Function

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• Euge
In summary, the limit of a nonnegative continuous function is the value that the function approaches but never quite reaches at a point. It is calculated by evaluating the function at values approaching the given point from both sides, and if the values approach the same value, then that value is the limit. The limit of a nonnegative continuous function is important because it helps us understand the behavior of the function at a particular point and is crucial in many mathematical and scientific applications. It cannot be negative and is represented on the graph as the horizontal asymptote, which helps us understand the overall behavior and shape of the function.
Euge
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Let ##D_r\subset \mathbb{R}^2## be the disk of radius ##r## centered at the origin. If ##f : \mathbb{R}^2 \to [0,\infty)## is uniformly continuous such that ##\sup_{0 < r< \infty} \iint_{D_r} f(x,y)\, dx\, dy < \infty##, show that ##f(x,y) \to 0## as ##x^2 + y^2 \to \infty##.

jbergman and topsquark
Since ##f## is uniformly continuous for every ##\epsilon >0## there exists ##\delta_{\epsilon} > 0## s.t. ##||(x_1,y_1)-(x_2,y_2)||<\delta ## implies ##|f(x_1,y_1)-f(x_2,y_2)|<\epsilon##.

Suppose that ##f## does not go to zero as ##x^2+y^2## goes to infinity. Since ##f## is non negative this means there exists some ##\epsilon >0## and a sequence of points ##(x_i,y_i)## such that ##x_i^2+y_i^2\to \infty ## and ##f(x_i,y_i)> 2 \epsilon##. Restrict to a subsequence such that ##\sqrt{x_i^2+y_i^2} > 2\delta_\epsilon \sqrt{x_{i-1}^2+y_{i-1}^2}##. Then the ball of radius ##\delta_{\epsilon}## around each ##(x_i,y_i)## do not intersect. We will call these balls ##B_i##

furthermore, on each ##B_i##, we have ##f> \epsilon## at every point, since it can be no more than ##\epsilon## smaller than ##f(x_i,y_i)\geq 2\epsilon##. Hence ##\int \int_{B_i}f(x,y)dx dy \geq \pi \delta_\epsilon^2 \epsilon##.

pick a sequence ##r_k## such that ##r_k>\sqrt{x_k^2+y_k^2}+\delta_{\epsilon}##. Then ##D_{r_k}## contains ##B_1,..., B_k## entirely, and since ## f## is non negative,

$$\int\int_{D_{r_k}} f(x,y)dx dy \geq \sum_{i=}^k \int \int_{B_i} f(x,y)dx dx\geq k\pi \delta_{\epsilon} ^2\epsilon$$

the last expression is just some constant multiplied by ##k##, which goes to infinity as ##k## does. This contradicts the assumption that the supremum of the integral over all radii choices is finite. Hence ##f## must go to zero as ##x^2+y^2\to \infty##

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