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anonymous
 4 years ago
can i solve
\[\lim_{x \rightarrow \infty} (2x)/\sqrt(9x^2 +1)\]
without using l'hopitals rule?
anonymous
 4 years ago
can i solve \[\lim_{x \rightarrow \infty} (2x)/\sqrt(9x^2 +1)\] without using l'hopitals rule?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure you can do it with your eyeballs

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0numerator is like x, denominator is like 3x and \[\frac{x}{3x}=\frac{1}{3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i meant eyeballs. just by looking

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.3yeah take x^2 out of the square root \[(2x) / (x \sqrt{9+1/x^2})\] take x out from numerator \[x(2/x1) / (x \sqrt{9+1/x^2})\] \[(2/x1) / ( \sqrt{9+1/x^2})\] 1/x and 1/x^2 >0 as x> \(\infty\) so we've \[ 1/ \sqrt 9\] 1/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is ok, but seems like a lot of work

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks ash for explanation, satellite your wit is endless

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually i was quite serious. suppose you wanted \[\lim_{x\rightarrow \infty}\frac{2x^2+3x+1}{5x^2+x2}\] wouldn't you say \[\frac{2}{5}\] just like that?
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