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- Homework Statement
- Prove the limit exists.

- Relevant Equations
- ##\text{lim}_{x\rightarrow a}\quad f=L## means ##\forall \epsilon>0, \exists \quad \delta ## such that ##|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon##.

Prove that each of the limits exists or does not exist.

1. ##\text{lim}_{x\rightarrow 2}(x^2-1)=3##

##\text{lim}_{x\rightarrow 2}(x^2-1)=3## if ##\forall \epsilon>0, \exists \delta ## such that ##|x-2|<\delta \Rightarrow |f(x)-3|<\epsilon##.

\begin{align}&|x^2-1|=|x+1||x-1|\leq \epsilon\\

\Rightarrow &|x-1|\leq \frac{\epsilon}{|x+1|}\\

&\delta=\frac{\epsilon}{|x+1|}\end{align}

2. ##\text{lim}_{x\rightarrow 1}(2x-1)\neq 2##

##\text{lim}_{x\rightarrow 1}(2x-1)\neq 2## if the limit exists elsewhere than ##2##. This means for all ##\epsilon>0##, there exists a ##\delta## such that ##|x-1|<\delta## implies ##\text{lim}_{x\rightarrow 1}(2x-1)= L## for some ##L\neq 2##.

\begin{align}&|2x-1-1|=2|x-1|\leq \epsilon\Rightarrow |x-1|\leq \frac{\epsilon}{2}\\

&\delta=\frac{\epsilon}{|2|}\end{align}

3. ##\text{lim}_{x\rightarrow \infty} (1+\frac{1}{n+1})=1##

We use ##x\rightarrow \frac{1}{x}## to transform the limit.

\begin{align}&\text{lim}_{x\rightarrow \infty} (1+\frac{1}{n+1})\rightarrow \text{lim}_{x\rightarrow 0} (1+\frac{n}{n+1})=1\\

&|1+\frac{n}{n+1}-1| \leq |n| <\epsilon \Rightarrow |n|\leq \epsilon\\

&\delta=\epsilon\\

&\therefore \forall \epsilon>0, |n|<\epsilon \Rightarrow |1+\frac{n}{n+1}-1|<\epsilon\\

&\therefore \forall \epsilon>0, |n-\infty|<\epsilon \Rightarrow |1+\frac{1}{n+1}-1|<\epsilon\\

\end{align}

1. ##\text{lim}_{x\rightarrow 2}(x^2-1)=3##

##\text{lim}_{x\rightarrow 2}(x^2-1)=3## if ##\forall \epsilon>0, \exists \delta ## such that ##|x-2|<\delta \Rightarrow |f(x)-3|<\epsilon##.

\begin{align}&|x^2-1|=|x+1||x-1|\leq \epsilon\\

\Rightarrow &|x-1|\leq \frac{\epsilon}{|x+1|}\\

&\delta=\frac{\epsilon}{|x+1|}\end{align}

2. ##\text{lim}_{x\rightarrow 1}(2x-1)\neq 2##

##\text{lim}_{x\rightarrow 1}(2x-1)\neq 2## if the limit exists elsewhere than ##2##. This means for all ##\epsilon>0##, there exists a ##\delta## such that ##|x-1|<\delta## implies ##\text{lim}_{x\rightarrow 1}(2x-1)= L## for some ##L\neq 2##.

\begin{align}&|2x-1-1|=2|x-1|\leq \epsilon\Rightarrow |x-1|\leq \frac{\epsilon}{2}\\

&\delta=\frac{\epsilon}{|2|}\end{align}

3. ##\text{lim}_{x\rightarrow \infty} (1+\frac{1}{n+1})=1##

We use ##x\rightarrow \frac{1}{x}## to transform the limit.

\begin{align}&\text{lim}_{x\rightarrow \infty} (1+\frac{1}{n+1})\rightarrow \text{lim}_{x\rightarrow 0} (1+\frac{n}{n+1})=1\\

&|1+\frac{n}{n+1}-1| \leq |n| <\epsilon \Rightarrow |n|\leq \epsilon\\

&\delta=\epsilon\\

&\therefore \forall \epsilon>0, |n|<\epsilon \Rightarrow |1+\frac{n}{n+1}-1|<\epsilon\\

&\therefore \forall \epsilon>0, |n-\infty|<\epsilon \Rightarrow |1+\frac{1}{n+1}-1|<\epsilon\\

\end{align}