### Understanding Continuity and Uniform Continuity#### Problem 2:Recall that we proved $ f(x) = \sqrt{x} $ is continuous on its domain $[0, \infty)$. Explain why $ f(x) $ is uniformly continuous on $[a, b]$ for any $ b > a \geq 0 $.---In this problem, we investigate the continuity properties of the square root function $ f(x) = \sqrt{x} $. We already know that it is continuous over the entire domain from zero to infinity. However, we need to explore why this function exhibits uniform continuity over any closed interval $[a, b]$, where $ b $ is greater than $ a $ and both are non-negative. **Definitions:**1. **Continuity**: A function $ f $ is continuous at a point $ x = c $ if: \[ \lim_{{x \to c}} f(x) = f(c) \]2. **Uniform Continuity**: A function $ f $ is uniformly continuous on a set $ S $ if for every $ \epsilon > 0 $, there exists a $ \delta > 0 $ such that for all $ x, y \in S $, if $ |x - y| < \delta $, then $ |f(x) - f(y)| < \epsilon $.**Explanation:**The key difference between continuity and uniform continuity is how the choice of $ \delta $ is determined. In continuity, $ \delta $ can depend on the specific point in the domain, while in uniform continuity, a single $ \delta $ must work for any point within the entire specified interval.For $ f(x) = \sqrt{x} $, on the interval $[a, b]$, the function does not have steep gradients or discontinuities, ensuring that small changes in $ x $ yield small changes in $ f(x) $. This consistent behavior across the interval is why the function is uniformly continuous.

Answered: 2. Recall that we proved f(z) = VE is…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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### Understanding Continuity and Uniform Continuity

#### Problem 2:
Recall that we proved $ f(x) = \sqrt{x} $ is continuous on its domain $[0, \infty)$. Explain why $ f(x) $ is uniformly continuous on $[a, b]$ for any $ b > a \geq 0 $.

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In this problem, we investigate the continuity properties of the square root function $ f(x) = \sqrt{x} $. We already know that it is continuous over the entire domain from zero to infinity. However, we need to explore why this function exhibits uniform continuity over any closed interval $[a, b]$, where $ b $ is greater than $ a $ and both are non-negative.

**Definitions:**

1. **Continuity**: A function $ f $ is continuous at a point $ x = c $ if:
\[
\lim_{{x \to c}} f(x) = f(c)
\]

2. **Uniform Continuity**: A function $ f $ is uniformly continuous on a set $ S $ if for every $ \epsilon > 0 $, there exists a $ \delta > 0 $ such that for all $ x, y \in S $, if $ |x - y| < \delta $, then $ |f(x) - f(y)| < \epsilon $.

**Explanation:**

The key difference between continuity and uniform continuity is how the choice of $ \delta $ is determined. In continuity, $ \delta $ can depend on the specific point in the domain, while in uniform continuity, a single $ \delta $ must work for any point within the entire specified interval.

For $ f(x) = \sqrt{x} $, on the interval $[a, b]$, the function does not have steep gradients or discontinuities, ensuring that small changes in $ x $ yield small changes in $ f(x) $. This consistent behavior across the interval is why the function is uniformly continuous.

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$d e f i n i t i o n : f o r a l l ε > 0 t h e r e e x i s t s a δ > 0, s u c h t h a t f o r a l l x, y \in X, | x - y | < δ \Rightarrow | f (x) - f (y) | < ε .$

Given $ε$ > 0 ,pick δ = $ε^{2}$ .

$F i r s t n o t e t h a t | \sqrt{x} - \sqrt{y} | \leq | \sqrt x + \sqrt y | .$

$H e n c e i f | x - y | < δ = ε^{2},$

$t h e n w e h a v e | \sqrt x - \sqrt y |^{2} \leq | \sqrt x - \sqrt y | | \sqrt x + \sqrt y | = | x - y | < ε^{2},$

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