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$\delta =$ \delta

In class practice

Proof Outline
Let $\epsilon >0$
Fix $\delta >0$ (Scratch work)
Let $x\in S$
Assume $0<|x-a|<\delta$
Reverse steps in scratch work.
Conclude $|f(x)-L|<\epsilon$

Prove that $\underset{x\to 4}{lim}2x+3=11$

Scratch work

Proof. Let $\epsilon >0$. Fix $\delta =\frac{\epsilon }{2}$. Let $x\in \mathbb{R}$. Assume that $0<|x-4|<\delta$. Then $0<|x-4|<\frac{\epsilon }{2}$. Thus $2|x-4|<\epsilon$. Hence $|2x-8|<\epsilon$. Thus $|(2x+3)-11|<\epsilon$. $◻$

Prove that $\underset{x\to 2}{lim}\frac{7x+4}{4x+1}=2$

Scratch work

Avoid $-\frac{1}{4}$ so that you're not where the function is undefined.

Require $|x-2|<1$ $x\in (1,3)$. $5<|4x+1|<13$.
$\frac{|x-2|}{5}<\epsilon$. $|x-2|<5\epsilon$. $\delta =min(1,5\epsilon )$

Proof.
Let $\epsilon >0$. Fix $\delta =min(1,5\epsilon )$. Let $x\in \mathbb{R}-\{-\frac{1}{4}\}$. Assume $0<|x-2|<\delta$. Then $|x-2|<min(1,5\epsilon )$. Thus $|x-2|<1$ and $|x-2|<5\epsilon$. Then $x\in (1,3)$. Then $5<|4x+1|<13$. Thus $\frac{|x-2|}{|4x+1|}<\frac{|x-2|}{5}$. Then $\frac{|x-2|}{|4x+1|}<\frac{5\epsilon }{5}$ since $|x-2|<5\epsilon$.

Thus

$◻$

$\underset{x\to 5}{lim}{x}^2+3x+3=43$

HW 35

Ben Finch

E 35.1
Prove that $\underset{x\to 4}{lim}2x+3=11$

Scratch work

Proof. Let $\epsilon >0$. Fix $\delta =\frac{\epsilon }{2}$. Let $x\in \mathbb{R}$. Assume that $0<|x-4|<\delta$. Then $0<|x-4|<\frac{\epsilon }{2}$. Thus $2|x-4|<\epsilon$. Hence $|2x-8|<\epsilon$. Thus $|(2x+3)-11|<\epsilon$. $◻$

E 35.2
Proof.
We want to show that $\underset{x\to a}{lim}cx+d=ca+d$.

Given an $\epsilon >0$, we need to find $\delta >0$ such that if $0<|x-a|<\delta$, then:

$|cx+d-(ca+d)|<\epsilon$

Simplifying the expression:

$|cx+d-ca-d|=|c(x-a)|$

Case 1: $c=0$

For $c=0$, the expression $|c(x-a)|$ becomes 0, and the condition $|0\cdot (x-a)|<\epsilon$ is trivially true for any $x$ since $|0\cdot (x-a)|=0$. Thus, we can choose any $\delta >0$, and the condition will be satisfied.

Case 2: $c\ne 0$

We need to find a $\delta$ such that $|c||x-a|<\epsilon$.

We can choose $\delta$ as follows:

$\delta =\frac{\epsilon }{|c|}$

Now, whenever $0<|x-a|<\delta$, we have:

$|c||x-a|<|c|\delta =|c|\cdot \frac{\epsilon }{|c|}=\epsilon$

Therefore, for every $\epsilon >0$, we have found a $\delta >0$ that satisfies the definition of the limit. $◻$

E 35.3
Prove that $\underset{x\to 5}{lim}{x}^2+3x+3=43$

Scratch work:

Require $|x-5|<1$, so $x\in (4,6)$. Then $x+8\in (12,14)$ and hence $12<x+8<14$. So $|(x-5)(x+8)|<|x-5|\cdot 14<\epsilon$. Then $|x-5|<\frac{\epsilon }{14}$. So choose $\delta =min(1,\frac{\epsilon }{14})$.

Proof.
Let $\epsilon >0$. Choose $\delta =min(1,\frac{\epsilon }{14})$. Let $x$ be any real number such that $0<|x-5|<\delta$.

1. If $0<|x-5|<1$, then $x$ is in the interval $(4,6)$. Therefore, $12<x+8<14$.

2. Now, $|(x+8)(x-5)|=|x-5|\cdot |x+8|<|x-5|\cdot 14<\frac{\epsilon }{14}\cdot 14=\epsilon$.

Hence, $\underset{x\to 5}{lim}{x}^2+3x+3=43$.

E 35.4
Prove that $\underset{x\to 2}{lim}\frac{7x+4}{4x+1}=2$

Scratch work

Avoid $-\frac{1}{4}$ so that you're not where the function is undefined.

Require $|x-2|<1$ $x\in (1,3)$. $5<|4x+1|<13$.
$\frac{|x-2|}{5}<\epsilon$. $|x-2|<5\epsilon$. $\delta =min(1,5\epsilon )$

Proof.
Let $\epsilon >0$. Fix $\delta =min(1,5\epsilon )$. Let $x\in \mathbb{R}-\{-\frac{1}{4}\}$. Assume $0<|x-2|<\delta$. Then $|x-2|<min(1,5\epsilon )$. Thus $|x-2|<1$ and $|x-2|<5\epsilon$. Then $x\in (1,3)$. Then $5<|4x+1|<13$. Thus $\frac{|x-2|}{|4x+1|}<\frac{|x-2|}{5}$. Then $\frac{|x-2|}{|4x+1|}<\frac{5\epsilon }{5}$ since $|x-2|<5\epsilon$. $◻$

E 35.5
Prove that $\underset{x\to 3}{lim}{x}^3+x^2+2=38$.

Scratch Work

Note that $x^3+x^2-36$ can be rewritten as $x^2(x+1)-36$. Now, consider $x\in (2,4)$, to ensure that $x+1$ and $x^2$ do not vary too wildly.

So we have:

Now, we need to find a $\delta$ such that when $0<|x-3|<\delta$, it follows that $|x^3+x^2-36|<\epsilon$. Since $|x^3+x^2-36|<116$ in our chosen interval, we can make $116<\epsilon$ by choosing $\delta$ small enough.

Thus, we can set $\delta =min(1,\frac{\epsilon }{116})$.

Proof.
Let $\epsilon >0$. Fix $\delta =min(1,\frac{\epsilon }{116})$. Let $x\in \mathbb{R}$. Assume $0<|x-3|<\delta$. Then $|x-3|<min(1,\frac{\epsilon }{116})$. Thus $x\in (2,4)$ and $|x^3+x^2-36|<116<\epsilon$. Hence, $\underset{x\to 3}{lim}{x}^3+x^2+2=38$. $◻$