Question

find the limit x approaches infinity, x-1/2x^2+3

Answer 1

i used the direct substitution and i got \[-\frac{ 1 }{ 3 }\] for the right side and \[\frac{ 1 }{ 3}\] for the left side. Is that right?

Answer 2

sorry, it is x approaches infinity.

Answer 3

oh

Answer 4

can you use l'Hopital's rule ?

Answer 5

divide the numerator and denominator by x \[\lim_{x\to\infty}\frac{x-1}{2x^2+3}\leadsto\frac\infty\infty\\
\lim_{x\to\infty}\frac{(x-1)/x}{(2x^2+3)/x}\]

Answer 6

@UnkleRhaukus it tends to zero

Answer 7

but we haven't explained to get there yet

Answer 8

you should tell him that the changes in denom are much greater than in nume then the whole of expression will approach zero..

Answer 9

yeah that's a different method ,
the powers of x in the denominator are greater than the powers of x in the numberator

Answer 10

\[\lim_{x\to\infty}\frac{x-1}{2x^2+3}\\=\lim_{x\to\infty}\frac{O(x)}{O(x^2)}\\=\lim_{x\to\infty}O(x^{-1})\]

Answer 11

yeah, actually and it would be zero as x goes to infinity..

Answer 12

I like l'Hopital's relue method best
\[\lim_{x\to\infty}\frac{x-1}{2x^2+3}\leadsto\frac\infty\infty\\
\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{\frac{\mathrm d}{dx}(x-1)}{\frac{\mathrm d}{dx}(2x^2+3)}\\=\lim_{x\to\infty}\frac{1}{4x}\\=\]

Answer 13

*rule

Answer 14

it is a bad theory

Answer 15

bad? how so?

Answer 16

it kills your creativity.

Answer 17

didn't kill mine

Answer 18

it doesn't even work every time

Answer 19

lol L'hops is to limits what FTC is to riemann sums. good to use L'hops when you're allowed to

Answer 20

it is okay only when you are going to make sure of what you've got by direct calculations..

Answer 21

taylor series of top and bottom would be more creative (save degrees of a fraction logic)

Answer 22

well sincethe greates power of numerator is less than the greatest power of denominator i would say limit is zero ( without thinking ) hmmm idk if i have to think on this xD

Answer 23

that was already said 2-3 times in above replies :P

Answer 24

actually ...

Answer 25

the problem is that how @dzie could ever get some nonzero answers.

Answer 26

thank you all.

Answer 27

but does that mean this cannot be solved using direct substitution?

Answer 28

i need more explanation please on why we need to divide the numerator and the denominator by x or why we need to use L'hopital rule

Answer 29

direct substitution leads to infinity / infinity, which is an indeterminate form,
so some technique must be applied to simplify the expression before the substitution is made

Answer 30

but why is it \[\frac{ \infty }{ \infty }\] this was what i did \[\frac{ \infty -1 }{ 2(\infty)^{2}+3}\]

Answer 31

\[\infty-1=\infty\\
2\times\infty^2+3=\infty\]

Answer 32

okay. thanks