Delta-epsilon proof of $\lim_{x\rightarrow\infty} \frac{x}{x+1} = 1$

Question

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: $$\lim_{x\rightarrow\infty} \frac{x}{x+1} = 1$$

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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 $\omega(\epsilon) = \frac{1}{\epsilon} - 1$

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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.

Answer 1

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, $$\lim_{x\rightarrow \infty} \frac{x}{x+1}=1$$