Question

use the precise definition of a limit to prove lim(x→ - 5)(x + 5)^4=0. specify a relationship between ε and δ that guarantees the limit exists. (hint: use the identity √4{x^4}=|x|). state the steps for proving that lim(x→a)f(x)=l. let ε be an arbitrary positive number. use the inequality to find a condition of the form where δ depends only on the value of ε. then, for any ε>0, assume and use the relationship between ε and δ to prove that

Explanation:

Step1: Recall the precise definition of a limit

The precise definition of $\lim_{x
ightarrow a}f(x) = L$ is: for every $\epsilon>0$, there exists a $\delta > 0$ such that if $0<|x - a|<\delta$, then $|f(x)-L|<\epsilon$. We want to prove $\lim_{x
ightarrow - 5}(x + 5)^4=0$. Let $f(x)=(x + 5)^4$, $a=-5$ and $L = 0$.

Step2: Set up the inequality

We need to find $\delta$ such that if $0<|x-(-5)|=|x + 5|<\delta$, then $|(x + 5)^4-0|=|(x + 5)^4|<\epsilon$.

Step3: Solve for $\delta$ in terms of $\epsilon$

Let $\delta=\sqrt[4]{\epsilon}$. Then, if $0<|x+5|<\delta=\sqrt[4]{\epsilon}$, we have $|(x + 5)^4|=(|x + 5|)^4<(\sqrt[4]{\epsilon})^4=\epsilon$.

Answer:

The relationship between $\epsilon$ and $\delta$ is $\delta=\sqrt[4]{\epsilon}$