1
$\begingroup$

Let's say if I know that $V \gg x^2$,

I know that for example:

$x^2 +V\sim V$,

however, if I have

$x^2 - V$,

can I still say that this equals

$x^2 - V\sim -V$?

Basic intuition tells me this must be so, based on taking numerical examples, i.e., if $1000 \gg 1$, then $1 - 1000 = -999 \sim -1000$ etc..., but I'm unsure if it is true in general.

Apologies in advanced for the basic question!

Thanks.

$\endgroup$
  • 4
    $\begingroup$ $\lim (x^2+V)/V = 1$ $\iff$ $\lim x^2/V = 0$ $\iff$ $\lim (x^2-V)/V = -1$. $\endgroup$ – dxiv Jan 3 '17 at 0:32
  • $\begingroup$ @dxiv What's that you say? :) Sorry, have not taken calculus yet. $\endgroup$ – Thomas Moore Jan 3 '17 at 0:40
  • $\begingroup$ is usually defined in terms of limits, so it helped if you explained what is your definition for it. $\endgroup$ – dxiv Jan 3 '17 at 0:55
  • $\begingroup$ If with ~ you mean ‘approximately equal to`, this is not a mathematical notion. $\endgroup$ – Bernard Jan 3 '17 at 1:30
  • $\begingroup$ $x^2+V-2V\sim( V-2V)=-V$ . $\endgroup$ – kingW3 Jan 3 '17 at 2:18

Your Answer

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

Browse other questions tagged or ask your own question.