# Delta-Epsilon Proof of an Reciprocal Function

I am trying to proof the following limit:
limx→11×2+1=12
\lim_{x \to 1}{\frac{1}{x^2+1}} = \frac{1}{2}

The question demanded this to be proven directly from the ϵ\epsilon – δ\delta definition of limit. However, I am struggling to find a value for δ\delta. Any insights?

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it is easier to prove that limx→1(x2+1)=2\lim_{x\to1}(x^2+1)=2 .
– Abdallah Hammam
Oct 20 at 21:17

@AbdallahHammam Of course, we can even plug x=1x=1 into the given expression, but I think we are supposed to do the ϵ−δ\epsilon-\delta proof for the given expression.
– Peter
Oct 20 at 21:21

@AbdallahHammam Judging from the fact that this question requires a delta – epsilon proof the be done, using limit laws may not be permitted.
– TommyX
Oct 20 at 21:22

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2

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|1×2+1−12|=|x2−12(x2+1)|=|x−1||x+1|2(x2+1)(∗)\left|\frac1{x^2+1}-\frac12\right|=\left|\frac{x^2-1}{2(x^2+1)}\right|=|x-1|\,\frac{|x+1|}{2(x^2+1)}\;\;\color{red}{(*)}

You have now to estimate the rightmost fraction on the right side, knowing that x\;x\; is going to be very close to 1\;1\;:

|x+1|2(x2+1)≤2.52=54\frac{|x+1|}{2(x^2+1)}\le\frac{2.5}{2}=\frac54

The above can be achieved, for example, by deciding that δ<12\;\delta<\frac12\; , say, and no matter what arbitrary ϵ>0\;\epsilon>0\; was chosen.

You now get

(∗)≤54δ\color{red}{(*)}\le\frac54\delta

…and now fill in details and end the proof.

I have also arrived at the step to estimate a value for
– TommyX
Oct 20 at 21:23

[continued from last comment] for the value of (). However I can’t seem to go from -1/2 < x + 1 < 1/2 to (). could you please elaborate on that step further? – TommyX Oct 20 at 21:25      +1 Clear and concise. – Dr. MV Oct 20 at 21:27      @tommyx Note that x2+1≥1x^2+1\ge 1. Then, for |x−1|≤1/2|x-1|\le 1/2, x+1≤5/2x+1\le 5/2 – Dr. MV Oct 20 at 21:28      I see. Thank you very much! – TommyX Oct 20 at 21:31 Let put f(x)=11+x2f(x)=\frac{1}{1+x^2}. Let ϵ>0\epsilon>0 given.

we look for δ\delta such that

|x−1|<δ⟹|f(x)−12|<ϵ.|x-1|<\delta \implies |f(x)-\frac{1}{2}|<\epsilon. but |f(x)−12|=|(1−x)(1+x)2(1+x2)|<|x−1||x+1|2|f(x)-\frac{1}{2}|=|\frac{(1-x)(1+x)}{2(1+x^2)}|<\frac{|x-1| |x+1|}{2}. as xx goes to 11, we can assume that xx satisfies |x−1|<2(=δ1)|x-1|<2(=\delta_1). thus |x+1|<4|x+1|<4. in th end |f(x)−12|<2|x−1||f(x)-\frac{1}{2}|<2|x-1|. So, we will choose δ\delta such that |x−1|<δ⟹2|x−1|<ϵ|x-1|<\delta \implies 2|x-1|<\epsilon or |x−1|<δ⟹|x−1|<ϵ2|x-1|<\delta \implies |x-1|<\frac{\epsilon}{2} in this case, we will be sure that |f(x)−12|<2|x−1|<ϵ|f(x)-\frac{1}{2}|<2|x-1|<\epsilon. we take δ=min(δ1,ϵ2)\delta=min(\delta_1,\frac{\epsilon}{2}) to satisfy both conditions |x−1|<δ1|x-1|<\delta_1 |x−1|<ϵ2|x-1|<\frac{\epsilon}{2}