anonymous
  • anonymous
can i solve \[\lim_{x \rightarrow \infty} (2-x)/\sqrt(9x^2 +1)\] without using l'hopitals rule?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
sure you can do it with your eyeballs
anonymous
  • anonymous
you mean algebra?
anonymous
  • anonymous
numerator is like -x, denominator is like 3x and \[\frac{-x}{3x}=-\frac{1}{3}\]

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anonymous
  • anonymous
no i meant eyeballs. just by looking
ash2326
  • ash2326
yeah take x^2 out of the square root \[(2-x) / (x \sqrt{9+1/x^2})\] take x out from numerator \[x(2/x-1) / (x \sqrt{9+1/x^2})\] \[(2/x-1) / ( \sqrt{9+1/x^2})\] 1/x and 1/x^2 ---->0 as x---> \(\infty\) so we've \[ -1/ \sqrt 9\] -1/3
anonymous
  • anonymous
that is ok, but seems like a lot of work
anonymous
  • anonymous
thanks ash for explanation, satellite your wit is endless
anonymous
  • anonymous
but not helpful
mathmate
  • mathmate
:)
anonymous
  • anonymous
actually i was quite serious. suppose you wanted \[\lim_{x\rightarrow \infty}\frac{2x^2+3x+1}{5x^2+x-2}\] wouldn't you say \[\frac{2}{5}\] just like that?

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