Suppose you have f(x)= $ \frac1{x^2+1}$ . We want to show that $\lim_{x\to {-1}} \frac1{x^2+1} = \frac12 $.
This is how I approached this issue
Suppose $ \lvert x+1 \rvert \lt \delta $ and $x \ne -1 $.
Then, $ \lvert f(x)-\frac12 \rvert = \lvert \frac1{x^2+1} -\frac12\rvert = \lvert\frac{-x^2 +1}{2(x^2 +1)}\rvert$. This can be simplified to: $ \lvert\frac{(1-x)(1+x)}{2(x^2 +1)}\rvert $. Therefore, let's assume that $ \lvert x+1 \rvert \lt 1 $.
So, $ -2 \lt x \lt 0 $. Hence, $ 0 \lt -x \lt 2 \iff 1 \lt -x+1 \lt 3 \iff \lvert -x+1 \rvert \lt 3$, and similarly, $ -2 \lt x \lt 0 \iff 0 \lt x^2 \lt 4 \iff 1 \lt x^2 +1 \lt 5 \iff \frac15 \lt \frac{1}{x^2 +1} \lt 1 \iff \lvert \frac1{x^2 +1} \rvert \lt 1. $
Hence, $ \lvert f(x)-\frac12 \rvert = \lvert\frac{(1-x)(1+x)}{2(x^2 +1)}\rvert \lt \frac{3. \lvert x+1 \rvert}{2.1} = \frac32 \lvert x+1 \rvert. $ To conclude, given any $ \epsilon \gt 0 $, let $ \delta$ = min (1, $ \frac{2 \epsilon}3) $.
Any pointers, mistakes or constructive criticism for this proof?
