# Epsilon-Delta proof: Use the formal definition to verify the indicated limit

I have a question. I have to prove the following limit by epsilon-delta argument. The limit is:

$\lim_{x \rightarrow 1}\frac{1}{x+1}=\frac{1}{2}$

I've used the formal definition $0<|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon$. I got (I'm starting with epsilon first).

$|f(x)-L|<\epsilon$

$|\frac{1}{x+1}-\frac{1}{2}|<\epsilon$

$|\frac{1-x}{2x+2}|<\epsilon$

And I'm stuck.

• Remember to vote and accept good answers received. It signals to all that the problem has been resolved. Moreover, it increases the probability that you will receive answers in the future. If a detail is unclear, then typically the author is happy to clarify the matter. Commented Nov 21, 2017 at 21:24

So the way to think about this is the actually try to "spell out" the definition with the proof. The symbolic definition is:

If $f$ is defined on a subset $D$ then

$$\lim_{x\mapsto c}f(x)=L \iff \left (\forall \varepsilon > 0,\,\exists \ \delta >0,\,\forall x\in D,\,0<|x-c|<\delta \ \Rightarrow \ |f(x)-L|<\varepsilon \right )$$

So, we know that $\varepsilon > 0$, thus we need to find a $\delta > 0$ so that $|x-c| < \delta$ implies that $|f(x)-L|<\varepsilon$. As i remember it, the tricky part is finding a form for the $\delta$ as a function of $\epsilon$. In this case:

$$|x-1| < \delta, \\ |\frac{1}{x+1} - \frac{1}{2}| = \frac{|x-1|}{2|x+1|} < \varepsilon \Leftrightarrow \frac{|x-1|}{|x+1|} < 2\varepsilon$$

so if i pick $\delta = 2\varepsilon$ we can state that

$$2\varepsilon > |x-1|$$

should imply that

$$2\varepsilon > \frac{|1-x|}{|x+1|}$$

This is true if there is a inequality

$$|x-1| \geq \frac{|1-x|}{|x+1|} \Leftrightarrow 1 \geq \frac{1}{|x+1|}\Leftrightarrow |x+1| \geq 1$$

And this inequality exists if $-2 \geq x \geq 0$, this implies that if we limit $\delta < 1$, then we know that $0 < x < 2$ and since this is within the frame of the inequality we know that $|x+1| \geq 1$ and thus the proof is done.

$|\dfrac{1}{x+1} -1/2| =$

$|\dfrac{2- x -1}{2(x+1)}| =$

$|\dfrac{1-x}{2(x+1)}|=$

$\dfrac{|x-1|}{|2(x+1)|} .$

Consider $x \in (0,2)$ I.e.

$1\lt (x+1) \lt 3$, and $|x-1| \lt 1$.

For a given $\epsilon \gt 0,$

choose $\delta = \min (1, 2 \epsilon)$ then

$|x-1|\lt \delta$ implies

$\dfrac{|x-1|}{2|x+1|} \lt$ $\dfrac{1}{2} \delta \le$ $\epsilon.$

We have to prove that $$\frac{1}{x+1} \rightarrow \frac{1}{2}, \quad x \rightarrow 1, \quad x \in \mathbb R, \quad x \not = 0,$$ by direct application of the definition. To that end, we first analyze the difference between our expression and the proposed limit. We have $$\left | \frac{1}{2} - \frac{1}{x+1} \right| = \left | \frac{x-1}{x+1} \right|.$$ Since we are interested in the limit where $x$ tends to $1$, there is no harm in concentrating on $x > 0$. It readily follows, that $$\left | \frac{1}{2} - \frac{1}{x+1} \right| \leq |x-1|, \quad x > 0$$ This completes the preliminary analysis of the problem. We are now ready to show $$\forall \epsilon>0 \: \exists \delta > 0 \: : \: |x-1| < \delta \: \Rightarrow \: \left | \frac{1}{2} - \frac{1}{x+1} \right| < \epsilon.$$ Let therefore $\epsilon > 0$ be given. Then by virtue of our analysis, we see that $$\delta = \min\{1,\epsilon\}$$ will suffice. In particular, $\delta \leq 1$ ensures that $x > 0$ such that our inequality applies.

This completes the proof.

Remark: Be wary of writing a sequence of isolated statements with no text or logical symbols indicating the relationship between them. Your reasoning can be perfectly correct, but you have failed in your responsibilty to demonstrate the correctness of your proof to the reader. A proof must convince the reader. It is better in the (Danish) words of Prof. Ebbe Thue Poulsen: "Et bevis er et overbevis".

The definition of $$\lim_{x\to1}\frac{1}{x+1}=\frac12$$ states $$\forall\epsilon\gt0,\exists\delta\gt0,\forall x\gt{-1},0\lt|x-1|\lt\delta\implies\bigg|\frac{1}{x+1}-\frac12\bigg|\lt\epsilon.$$ Now notice that $$\big|\frac{1}{x+1}-\frac12\big|=\big|\frac{2-(x+1)}{2(x+1)}\big|=\frac12\big|\frac{x-1}{x+1}\big|=\frac12\frac{|x-1|}{|x+1|}\lt\frac{\delta}{2|x+1|}$$. Thus, if we can find an upper bound $$U_{\delta}$$ in terms of $$\delta$$ such that $$\frac{1}{2|x+1|}\lt{U_{\delta}}$$, then we can let $$\epsilon=\delta{U_{\delta}}$$.

For this, we go back to $$0\lt|x-1|=|(x+1)-2|\lt\delta$$. The reverse triangle inequality implies that $$||x+1|-2|\leq|x-1|\lt\delta$$, which is equivalent to $$-\delta\lt|x+1|-2\lt\delta$$, which is equivalent to $$2-\delta\lt|x+1|\lt2+\delta$$, of course with the added caveat that $$|x+1|\neq0$$. Now, notice that, if we assume $$0\lt2-\delta$$, then $$2-\delta\lt|x+1|\lt2+\delta$$ implies $$\frac{1}{2+\delta}\lt\frac{1}{|x+1|}\lt\frac{1}{2-\delta}$$. This gives us that $$\frac{1}{2|x+1|}\lt{U_{\delta}}=\frac{1}{2(2-\delta)}$$. Thus $$\epsilon=\frac{\delta}{2(2-\delta)}$$, which is equivalent to $$4\epsilon-2\epsilon\delta=\delta$$, which is equivalent to $$\delta=\frac{4\epsilon}{2\epsilon+1}$$, which, for every $$\epsilon\gt0$$, does indeed satisfy that $$\delta\gt0$$.

Therefore, $$\forall\epsilon\gt0,\exists\delta=\frac{4\epsilon}{2\epsilon+1}\gt0,\forall{x\gt{-1}},0\lt|x-1|\lt\delta\implies\bigg|\frac{1}{x+1}-\frac12\bigg|\lt\epsilon.$$ Therefore, $$\lim_{x\to1}\frac{1}{x+1}=\frac12.$$ Q.E.D.