use the method in example 23.2 to prove the question. **Example 23.2.** Let $ f(x) $ be any linear function: $ f(x) = mx + b $ where $ m $ and $ b $ are fixed real numbers. (The domain of $ f $ is all of $\mathbb{R}$.) Let $ a $ be any real number. I claim\[\lim_{x \to a} f(x) = f(a).\](As we will see, this is equivalent to the property that $ f $ is "continuous" at $ x = a $.) Pick $ \epsilon > 0 $.(Scratch work: We need $|f(x) - f(a)| < \epsilon$ and this is equivalent to $|m(x - a)| < \epsilon$ which in turn (provided $ m \neq 0 $) is equivalent to $|x - a| < \frac{\epsilon}{|m|}$. If $ m = 0 $, then $|f(x) - f(a)| < \epsilon$ is automatic.)We proceed in cases.**Case I:** Suppose $ m = 0 $. Then set $\delta = 10^{100}$ (or any positive number you like). If $ 0 < |x - a| < \delta $, then $ f $ is defined at $ x $ and $|f(x) - f(a)| = |b - b| = 0 < \epsilon$. This proves $\lim_{x \to a} f(x) = f(a)$.**Case II:** Suppose $ m \neq 0 $. Set $\delta = \frac{\epsilon}{|m|}$. If $ 0 < |x - a| < \delta $, then $ f $ is defined at $ x $ and\[|f(x) - f(a)| = |mx + b - (ma + b)| = |m(x - a)| = |m||x - a| < |m|\delta = \epsilon.\]This proves $\lim_{x \to a} f(x) = f(a)$. **Using the $\epsilon - \delta$ Definition of Limits:****Problem Statement:**Let $ a > 0 $. Using just the $\epsilon - \delta$ definition of the limit of a function, prove that\[\lim_{x \to a} \sqrt{x} = \sqrt{a}.\]**Tip:** Use the inequality\[|\sqrt{x} - \sqrt{a}| = \frac{|x - a|}{\sqrt{x} + \sqrt{a}} \leq \frac{|x-a|}{\sqrt{a}}.\]**Explanation:**- The expression $|\sqrt{x} - \sqrt{a}|$ uses a rationalization trick, where the difference of square roots is expressed in terms of a fraction, making it easier to manage under limit calculations.- The inequality $\frac{|x-a|}{\sqrt{a}}$ shows an upper bound that can be employed in the $\epsilon - \delta$ definition, facilitating the proof of the limit.

Name: use the method in example 23.2 to prove the question. **Example 23.2.*
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Description: use the method in example 23.2 to prove the question. **Example 23.2.** Let $ f(x) $ be any linear function: $ f(x) = mx + b $ where $ m $ and $ b $ are fixed real numbers. (The domain of $ f $ is all of $\mathbb{R}$.) Let $ a $ be any

Answered: Let a >0. Using just the e – 6…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

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use the method in example 23.2 to prove the question.

**Example 23.2.** Let $ f(x) $ be any linear function: $ f(x) = mx + b $ where $ m $ and $ b $ are fixed real numbers. (The domain of $ f $ is all of $\mathbb{R}$.) Let $ a $ be any real number. I claim

\[
\lim_{x \to a} f(x) = f(a).
\]

(As we will see, this is equivalent to the property that $ f $ is "continuous" at $ x = a $.) Pick $ \epsilon > 0 $.

(Scratch work: We need $|f(x) - f(a)| < \epsilon$ and this is equivalent to $|m(x - a)| < \epsilon$ which in turn (provided $ m \neq 0 $) is equivalent to $|x - a| < \frac{\epsilon}{|m|}$. If $ m = 0 $, then $|f(x) - f(a)| < \epsilon$ is automatic.)

We proceed in cases.

**Case I:** Suppose $ m = 0 $. Then set $\delta = 10^{100}$ (or any positive number you like). If $ 0 < |x - a| < \delta $, then $ f $ is defined at $ x $ and $|f(x) - f(a)| = |b - b| = 0 < \epsilon$. This proves $\lim_{x \to a} f(x) = f(a)$.

**Case II:** Suppose $ m \neq 0 $. Set $\delta = \frac{\epsilon}{|m|}$. If $ 0 < |x - a| < \delta $, then $ f $ is defined at $ x $ and

\[
|f(x) - f(a)| = |mx + b - (ma + b)| = |m(x - a)| = |m||x - a| < |m|\delta = \epsilon.
\]

This proves $\lim_{x \to a} f(x) = f(a)$.

**Using the $\epsilon - \delta$ Definition of Limits:**

**Problem Statement:**
Let $ a > 0 $. Using just the $\epsilon - \delta$ definition of the limit of a function, prove that

\[
\lim_{x \to a} \sqrt{x} = \sqrt{a}.
\]

**Tip:** Use the inequality

\[
|\sqrt{x} - \sqrt{a}| = \frac{|x - a|}{\sqrt{x} + \sqrt{a}} \leq \frac{|x-a|}{\sqrt{a}}.
\]

**Explanation:**
- The expression $|\sqrt{x} - \sqrt{a}|$ uses a rationalization trick, where the difference of square roots is expressed in terms of a fraction, making it easier to manage under limit calculations.
- The inequality $\frac{|x-a|}{\sqrt{a}}$ shows an upper bound that can be employed in the $\epsilon - \delta$ definition, facilitating the proof of the limit.

Expert Solution

Step 1

The limit of function of f is given by,

Let I be an open interval containing c, and let f be the function defined on I. The limit of f(x), as x approaches c, is L , denoted by

$\lim_{x \to c} f (x) = L$

that is for any $ε$ >0, there exists $δ$ >0 such that for all $x \neq c$ , if $| x - c | < δ,$ then $| f (x) - L | < ε$

Which also be stated as,

For every $ε > 0$ there exists $δ > 0$ such that $0 < | x - a | < δ$ implies that $| f (x) - f (a) | < ε$

Here $f (x) = \sqrt{x} f (a) = \sqrt{a}$

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