Let a >0. Using just the e – 6 definition of the limit of a function, prove that lim V = Va. |r – a| Va Tip: Use |Vī – va]= |I – | %3D
Percentage
A percentage is a number indicated as a fraction of 100. It is a dimensionless number often expressed using the symbol %.
Algebraic Expressions
In mathematics, an algebraic expression consists of constant(s), variable(s), and mathematical operators. It is made up of terms.
Numbers
Numbers are some measures used for counting. They can be compared one with another to know its position in the number line and determine which one is greater or lesser than the other.
Subtraction
Before we begin to understand the subtraction of algebraic expressions, we need to list out a few things that form the basis of algebra.
Addition
Before we begin to understand the addition of algebraic expressions, we need to list out a few things that form the basis of algebra.
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 \(\lim_{x \to a} f(x) = f(a)\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fba5a3306-2973-4825-a0c7-f3a4b662c857%2F395aeaaa-9fc9-4043-8715-d69d4f915db9%2Fsuqjn1i_processed.png&w=3840&q=75)
![**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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fba5a3306-2973-4825-a0c7-f3a4b662c857%2F395aeaaa-9fc9-4043-8715-d69d4f915db9%2Ft0wndj8_processed.png&w=3840&q=75)

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
that is for any >0, there exists >0 such that for all , if then
Which also be stated as,
For every there exists such that implies that
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