7. -/6 points details my notes ask your teacher practice another prove the statement using the ε, δ…

7. -/6 points details my notes ask your teacher practice another prove the statement using the ε, δ definition of a limit. $lim_{x\rightarrow - 2}(x^{2}-1)=3$ given $epsilon>0$, we need $delta$ such that if $0 < |x-(-2)|<delta$, then $|(x^{2}-1)-3|$ or upon simplifying we need $|x^{2}-4|<epsilon$ whenever $0 < |x + 2|<delta$. notice that if $|x + 2|<1$, then $-1<x + 2<1rightarrow - 5<x-2<-3rightarrow|x - 2|<$ so take $delta =$. then $0 < |x + 2|<deltarightarrow|x - 2|<5$ and $|x + 2|<epsilon/5$, so $|(x^{2}-1)-3|=|(x + 2)(x - 2)|=|x + 2||x - 2|<(epsilon/5)(5)=$ thus, by the definition of a limit, $lim_{x\rightarrow - 2}(x^{2}-1)=$

7. -/6 points details my notes ask your teacher practice another prove the statement using the ε, δ definition of a limit. $lim_{x\rightarrow - 2}(x^{2}-1)=3$ given $epsilon>0$, we need $delta$ such that if $0 < |x-(-2)|<delta$, then $|(x^{2}-1)-3|$ or upon simplifying we need $|x^{2}-4|<epsilon$ whenever $0 < |x + 2|<delta$. notice that if $|x + 2|<1$, then $-1<x + 2<1rightarrow - 5<x-2<-3rightarrow|x - 2|<$ so take $delta =$. then $0 < |x + 2|<deltarightarrow|x - 2|<5$ and $|x + 2|<epsilon/5$, so $|(x^{2}-1)-3|=|(x + 2)(x - 2)|=|x + 2||x - 2|<(epsilon/5)(5)=$ thus, by the definition of a limit, $lim_{x\rightarrow - 2}(x^{2}-1)=$

Answer

Explanation:

Step1: Recall the $\epsilon - \delta$ definition of limit

We want to show that for every $\epsilon>0$, there exists a $\delta > 0$ such that if $0<|x - (- 2)|<\delta$, then $|(x^{2}-1)-3|<\epsilon$. First, simplify $|(x^{2}-1)-3|$: [|(x^{2}-1)-3|=|x^{2}-4|=|(x + 2)(x - 2)|]

Step2: Bound $|x - 2|$ in terms of $|x+2|$

Suppose $|x + 2|<1$. Then $-1<x + 2<1$, which implies $-5<x-2<-3$, so $|x - 2|<5$.

Step3: Choose $\delta$

We want $|(x + 2)(x - 2)|<\epsilon$. Since $|x - 2|<5$ when $|x + 2|<1$, we choose $\delta=\min\left{1,\frac{\epsilon}{5}\right}$.

Step4: Prove the limit

If $0<|x + 2|<\delta$, then $|(x^{2}-1)-3|=|(x + 2)(x - 2)|$. Since $|x + 2|<\delta\leqslant\frac{\epsilon}{5}$ and $|x - 2|<5$ (because $|x + 2|<\delta\leqslant1$), we have $|(x + 2)(x - 2)|<\frac{\epsilon}{5}\times5=\epsilon$.

Answer:

We choose $\delta=\min\left{1,\frac{\epsilon}{5}\right}$ to prove $\lim_{x\rightarrow - 2}(x^{2}-1)=3$ using the $\epsilon-\delta$ definition of a limit.