Question

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

Answer 1

sure you can do it with your eyeballs

Answer 2

you mean algebra?

Answer 3

numerator is like -x, denominator is like 3x
and
\[\frac{-x}{3x}=-\frac{1}{3}\]

Answer 4

no i meant eyeballs. just by looking

Answer 5

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

Answer 6

that is ok, but seems like a lot of work

Answer 7

thanks ash for explanation, satellite your wit is endless

Answer 8

but not helpful

Answer 9

:)

Answer 10

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?