1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $x$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ True that. Would have been it simpler! Thanks for pointing it out! :) $\endgroup$ – jeff123 Jun 4 at 2:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.