# Limits: Epsilon and Delta questions on proving limits

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?

Looks good to me. But one can simplify it further by noticing that $$\displaystyle\left|\frac{1}{x^2+1}\right|\leq 1$$ for any $$x$$.