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.

  • $\begingroup$ This looks great! Is there a particular part you're especially concerned about? $\endgroup$ Oct 14, 2015 at 20:47
  • $\begingroup$ Welcome to MSE. This is a really well-composed first post. Keep it up. $\endgroup$
    – zahbaz
    Oct 14, 2015 at 20:49
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    $\begingroup$ You have made an incorrect use of the symbol $\iff$. Instead use $=$. $\endgroup$ Oct 14, 2015 at 20:58
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    $\begingroup$ In $A\iff B$, $A$ and $B$ must be statements. A statement is a sentence that is true or false. Is $\vert x / (x+1) - 1\rvert$ true ? But "$\lvert x/(x+1) - 1\rvert = \lvert -1/(x+1) \rvert$ for all real $x\neq -1$" is a true statement. $\endgroup$ Oct 14, 2015 at 21:11
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    $\begingroup$ @Skurmedel Well, yes, the tradition is to do the "scratch work" on the side, and write the nice, formal proof in the opposite direction. I will include a more traditional proof below. $\endgroup$ Oct 14, 2015 at 21:14

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

  • $\begingroup$ Ah, ok, I see. Yeah so you need both, and this is the final product. I guess that's a bit more on the point. It's weird that my American book about calculus does it this way sometimes, but my native language books go the other way. If it's just different conventions or pedagogy I don't know. Probably a pedagogic decision I reckon. $\endgroup$
    – Skurmedel
    Oct 15, 2015 at 4:30

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