prove the statement using the \\(\\epsilon, \\delta\\) definition of a limit.\\(\\lim_{x\\to…

prove the statement using the \\(\\epsilon, \\delta\\) definition of a limit.\\(\\lim_{x\\to - 2}(x^{2}-1)=3\\)given \\(\\epsilon>0\\), we need \\(\\delta\\) such that if \\(0 <|x - (-2)|<\\delta\\), then \\(|(x^{2}-1)-3|<\\epsilon\\).\\(|x + 2|<1\\), then \\(-1 < x+2 < 1\\rightarrow - 5 < x - 2 < - 3\\rightarrow|x - 2|<5\\).\\(|(x^{2}-1)-3|=|(x + 2)(x - 2)|=|x + 2||x - 2|<(\\epsilon/5)(5)=\\epsilon\\). thus, by the definition of a limit, \\(\\lim_{x\\to - 2}(x^{2}-1)=3\\). so take \\(\\delta =\\) select \\(\\vee\\) upon simplifying we need \\(|x^{2}-4|<\\epsilon\\) whenever \\(0 <|x + 2|<\\delta\\). notice that if \\(0 <|x + 2|<\\delta\\rightarrow|x - 2|<5\\) and \\(|x + 2|<\\epsilon/5\\), so \\(|x + 2|<1\\rightarrow - 1 < x+2 < 1\\rightarrow - 5 < x - 2 < - 3\\rightarrow|x - 2|<5\\). then \\(0 <|x + 2|<\\delta\\rightarrow|x - 2|<5\\) and \\(|x + 2|<\\epsilon/5\\).

prove the statement using the \\(\\epsilon, \\delta\\) definition of a limit.\\(\\lim_{x\\to - 2}(x^{2}-1)=3\\)given \\(\\epsilon>0\\), we need \\(\\delta\\) such that if \\(0 <|x - (-2)|<\\delta\\), then \\(|(x^{2}-1)-3|<\\epsilon\\).\\(|x + 2|<1\\), then \\(-1 < x+2 < 1\\rightarrow - 5 < x - 2 < - 3\\rightarrow|x - 2|<5\\).\\(|(x^{2}-1)-3|=|(x + 2)(x - 2)|=|x + 2||x - 2|<(\\epsilon/5)(5)=\\epsilon\\). thus, by the definition of a limit, \\(\\lim_{x\\to - 2}(x^{2}-1)=3\\). so take \\(\\delta =\\) select \\(\\vee\\) upon simplifying we need \\(|x^{2}-4|<\\epsilon\\) whenever \\(0 <|x + 2|<\\delta\\). notice that if \\(0 <|x + 2|<\\delta\\rightarrow|x - 2|<5\\) and \\(|x + 2|<\\epsilon/5\\), so \\(|x + 2|<1\\rightarrow - 1 < x+2 < 1\\rightarrow - 5 < x - 2 < - 3\\rightarrow|x - 2|<5\\). then \\(0 <|x + 2|<\\delta\\rightarrow|x - 2|<5\\) and \\(|x + 2|<\\epsilon/5\\).

Answer

Explanation:

Step1: Start with the limit - definition

We want to prove $\lim_{x\rightarrow - 2}(x^{2}-1)=3$ using the $\epsilon-\delta$ definition. The $\epsilon-\delta$ definition of a limit states that $\lim_{x\rightarrow a}f(x)=L$ if for every $\epsilon>0$, there exists a $\delta > 0$ such that if $0<|x - a|<\delta$, then $|f(x)-L|<\epsilon$. Here, $a=-2$, $f(x)=x^{2}-1$, and $L = 3$. So we need to show that if $0<|x+2|<\delta$, then $|(x^{2}-1)-3|=|x^{2}-4|<\epsilon$.

Step2: Factor the expression

Factor $|x^{2}-4|$ as $|(x + 2)(x - 2)|=|x + 2||x - 2|$.

Step3: Bound $|x - 2|$

If $|x+2|<1$, then $-1<x + 2<1$. Subtracting 4 from all parts of the inequality gives $-5<x - 2<-3$, so $|x - 2|<5$.

Step4: Relate $\epsilon$ and $\delta$

We have $|x^{2}-4|=|x + 2||x - 2|$. We want $|x^{2}-4|<\epsilon$. Since $|x - 2|<5$ when $|x+2|<1$, if we also have $|x + 2|<\frac{\epsilon}{5}$, then $|x^{2}-4|=|x + 2||x - 2|<5\times\frac{\epsilon}{5}=\epsilon$. So we take $\delta=\min\left{1,\frac{\epsilon}{5}\right}$.

Answer:

$\delta=\min\left{1,\frac{\epsilon}{5}\right}$