Question

consider the limit $lim_{x
ightarrow1}(4x + 9)=13$. when checking the epsilon - delta definition of the limit, if $epsilon$ is given as 0.4, what is the largest that $delta$ can be? $delta=$ finish the definition of the epsilon - delta definition of the limit $lim_{x
ightarrow a}f(x)=l$: given $epsilon>0$, $delta > 0$ so that if then

Explanation:

Step1: Recall epsilon - delta definition

The definition of $\lim_{x
ightarrow a}f(x)=L$ is: Given $\epsilon>0$, there exists $\delta > 0$ such that if $0<|x - a|<\delta$, then $|f(x)-L|<\epsilon$. Here $f(x)=4x + 9$, $a = 1$, $L=13$ and $\epsilon=0.4$.

Step2: Set up the inequality

We want $|(4x + 9)-13|<\epsilon$. Simplify $|(4x + 9)-13|$ to $|4x - 4|=4|x - 1|$. So we have $4|x - 1|<\epsilon$.

Step3: Solve for $\delta$

Since $4|x - 1|<\epsilon$ and we know that $|x - 1|<\delta$, we can solve the inequality $4|x - 1|<\epsilon$ for $|x - 1|$. Divide both sides of $4|x - 1|<0.4$ by 4, we get $|x - 1|<\frac{0.4}{4}$.

Answer:

$0.1$