TITLE: Delta-epsilon proof of $\lim_{x\rightarrow\infty} \frac{x}{x+1} = 1$
QUESTION [6 upvotes]: I have an exercise where I'm supposed to show, by delta-epsilon proof that $\frac{x}{x+1}$ tends to 1 as $x$ goes to positive infinity.
In our faculty and literature, for limits at infinity we usually call $\delta$ small omega ($\omega$) instead. So the definition I use is the following:
$$x > \omega \Rightarrow |f(x)-A|\leq\epsilon$$
where
$$x > 0,\; \omega(\epsilon),\; \epsilon > 0$$
So pretty standard definition.
Now here's my attempted proof of:
$$\lim_{x\rightarrow\infty} \frac{x}{x+1} = 1$$

We have
$$\left|\frac{x}{x+1}-1\right|\Leftrightarrow \left|-\frac{1}{x+1}\right|$$
Also for positive $x$, $x + 1 > 0$ so:
$$\frac{1}{x+1} \leq \epsilon$$
Which (again with assumption $x > 0$) gives:
$$\frac{1}{\epsilon} - 1 \leq x$$
so we can use $\omega(\epsilon) = \frac{1}{\epsilon} - 1$

I am struggling somewhat in real analysis at the moment, so I have very low confidence that I'm not missing something important. It would be greatly appreciated if someone could take a look at my proof and give feedback.

REPLY [2 votes]: Proof going in the "opposite direction" as Skurmedel's... Skurmedel's work resembles very good "scratch work", whose goal is to figure out the appropriate $\delta(\epsilon)$.  One usually does the work he has done, then turns in a final proof like this:
Let $0<\epsilon<1$.  Put $\delta(\epsilon):=\frac{1}{\epsilon}-1$  Then $$\left|\frac{x}{x+1}-1\right|=\left|\frac{-1}{x+1}\right|=\frac{1}{x+1}$$ for $x>-1$ (and, in particular, for $x>0$).  Then $$x>\delta(\epsilon)=\frac{1}{\epsilon}-1$$ implies $$x+1>\frac{1}{\epsilon}$$ and hence, since $x+1>0$ for $x>0$, this last statement is equivalent to $$\epsilon >\frac{1}{x+1}(>0)$$ Therefore, $$\lim_{x\rightarrow \infty} \frac{x}{x+1}=1$$