Delta-Epsilon Proof of an Reciprocal Function

Question

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?

Answer 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 $\;x\;$ is going to be very close to $\;1\;$:

$$\frac{|x+1|}{2(x^2+1)}\le\frac{2.5}{2}=\frac54$$

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

$$\color{red}{(*)}\le\frac54\delta$$

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

Answer 2

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

Let $\epsilon>0$ given.

we look for $\delta$ such that

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

but

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

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

thus $|x+1|<4$.

in th end

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

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

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

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

to satisfy both conditions

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

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