I am trying to proof the following limit: $$ \lim_{x \to 1}{\frac{1}{x^2+1}} = \frac{1}{2} $$ The question demanded this to be proven directly from the $\epsilon$ - $\delta$ definition of limit. However, I am struggling to find a value for $\delta$. Any insights?

  • $\begingroup$ it is easier to prove that $\lim_{x\to1}(x^2+1)=2$ . $\endgroup$ – hamam_Abdallah Oct 20 '16 at 21:17
  • $\begingroup$ @AbdallahHammam Of course, we can even plug $x=1$ into the given expression, but I think we are supposed to do the $\epsilon-\delta$ proof for the given expression. $\endgroup$ – Peter Oct 20 '16 at 21:21
  • $\begingroup$ @AbdallahHammam Judging from the fact that this question requires a delta - epsilon proof the be done, using limit laws may not be permitted. $\endgroup$ – TommyX Oct 20 '16 at 21:22


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


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

You now get


...and now fill in details and end the proof.

  • $\begingroup$ I have also arrived at the step to estimate a value for $\endgroup$ – TommyX Oct 20 '16 at 21:23
  • $\begingroup$ [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? $\endgroup$ – TommyX Oct 20 '16 at 21:25
  • $\begingroup$ @tommyx Note that $x^2+1\ge 1$. Then, for $|x-1|\le 1/2$, $x+1\le 5/2$ $\endgroup$ – Mark Viola Oct 20 '16 at 21:28
  • $\begingroup$ I see. Thank you very much! $\endgroup$ – TommyX Oct 20 '16 at 21:31

Let put $f(x)=\frac{1}{1+x^2}$.

Let $\epsilon>0$ given.

we look for $\delta$ such that

$|x-1|<\delta \implies |f(x)-\frac{1}{2}|<\epsilon.$


$|f(x)-\frac{1}{2}|=|\frac{(1-x)(1+x)}{2(1+x^2)}|<\frac{|x-1| |x+1|}{2}$.

as $x$ goes to $1$, we can assume that $x$ satisfies $|x-1|<2(=\delta_1)$.

thus $|x+1|<4$.

in th end


So, we will choose $\delta$ such that

$|x-1|<\delta \implies 2|x-1|<\epsilon$ or $|x-1|<\delta \implies |x-1|<\frac{\epsilon}{2}$

in this case, we will be sure that


we take $\delta=min(\delta_1,\frac{\epsilon}{2})$

to satisfy both conditions

  1. $|x-1|<\delta_1$

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


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