Question

calculus 1 limits and derivatives chapter help. delta and epsilon proof​

Answer 1

Answer:

$\delta=\frac{\epsilon}{3}$

The proof is in the explanation.

Step-by-step explanation:

Compare $\lim_{x\rightarrow 7}(3x-2)=19$ to $\lim_{x\rightarrow a}f(x)=L$.

We want to find  $\delta$ such that whenever $0<|x-a|<\delta$ we have $|f(x)-L|<\epsilon$.

We want to find $\delta$ such that whenever $0<|x-7|<\delta$ we have $|3x-2-19|<\epsilon$.

$|3x-2-19|<\epsilon$

Simplify:

$|3x-21|<\epsilon$

Factor on left side inside the | |:

$|3(x-7)|<\epsilon$

|ab|=|a||b|:

$|3||x-7|<\epsilon$

|3|=3:

$3|x-7|<\epsilon$

Divide both sides by 3:

$|x-7|<\frac{\epsilon}{3}$

So we will need to choose $\delta=\frac{\epsilon}{3}$.

We want to prove there exists a $\delta$ such that whenever $0<|x-7|<\delta$ we have $|f(x)-L|<\epsilon$.

Proof:

Let $\epsilon>0$. We will choose $\delta=\frac{\epsilon}{3}$ such that $0<|x-7|<\delta$ will give us $|f(x)-L|<\epsilon$.

$|f(x)-L|=|(3x-2)-19|$

        $=|3x-21|$

        $=|3||x-7|$

        $=3|x-7|<3\delta=3\frac{\epsilon}{3}=\epsilon$.

Therefore, we have for all $\epsilon>0$, $\delta=\frac{\epsilon}{3}$ is the number that satisfies the definition.