This is the Question :
Given the following limits, find $M>0$ such that for every $x>M$ the expression are $\frac13$-close to their limit (in other words, find $M>0$ s.t. for every $x>M$: |f(x)-L|<\frac13$ for the following functions):
(a) $\lim_{x\to\infty}\frac{x^2+2x-3}{x^2-1}=1$
I'm not sure that i am solving this right. What I did was finding $M = 2/\epsilon$ using the limit defintion. Then I am asked to make sure that $\epsilon < \frac13$ so I did $\frac2M \le \frac13$ which gave me $M > 6$. So I am thinking that for the final answer I can choose $M = 6 + \frac2\epsilon$ and that will make sure that the function will stay between $1.333$ to $\frac23$.
My logic is correct or did I just write very weird and uncorrect stuff ? Thanks in advance !
Note that $$\frac{x^2+2x-3}{x^2-1}-1=\frac{2x-2}{x^2-1}=\frac2{x+1} $$ provided (which is needed anyway) $x\ne 1$. The desired condition is $\left|\frac2{x+1}\right|<\frac13$, which is equivalent to ($x>-1$ and $x\ne 1$ and $\frac2{x+1}<\frac13$) or ($x<-1$ and $\frac2{x+1}>\frac13$). Per problem statement we are interested only in positive $M$ and hence only in positive $x$. Therefore, we can drop the second branch and consider $\left|\frac2{x+1}\right|<\frac13$ as equivalent to $$ \frac2{x+1}<\frac13\iff 2<\frac13(x+1)\iff 6<x+1\iff 5<x.$$ This allows us to pick $$ M=5.$$ (Fortunately, this automatically gets rid of the problem at $x=1$). This is the smallest possible choice of $M$. As long as the problem statement does not require you to find the smallest possible $M$, any larger value will be just as correct, for example $M=42$.