Calculus 1 question... I need help solving limit problems as it approaches infinity

[skim milk]

sorry to ask again but can someone help me out?

the questions is lim as x->infinity of squareroot of (9X+2) / 4X+3

the square root is just for the numerator

i graphed it and it looks like it is approaching 0 as x->0 but i want to see how it is done

also what about the lim as x> - infinity? (negative infinity)

 

[Syringer]

Have you learned L'Hospital's rule?

 

[KingNothing]

/reminds himself not to care about your opinion on whether or not .999... = 1.

 

[CrazyPerson]

i am seeing stars

 

[GiLtY]

Originally posted by: Syringer
Have you learned L'Hospital's rule?

Yah, L'Hospital's Rule is the best way to go, but you'd have to know derivatives first..

 

[simms]

 

[skim milk]

no i don't think we use l'hopital's rule since the teacher skipped over it

the thing is i was late to class so i missed out on the lecture... that's why i am asking here... sorry :frown:

 

[wildwarren]

skipped over L'Hospital's? Seems funny....

 

[skim milk]

Originally posted by: simms
divide everything by x.

how do i deal with the square root in the numerator?

if i take the coefficients of the highest powers,,,, it would be square root of 9 over 4 which comes out to be 3/4 = .75 but the graph surely shows that it is approaching 0

 

[simms]

This should be right.. i have my calculus exam tommorow.

That's it.. i'm taking out the Ti-89.

 

[skim milk]

Originally posted by: simms
This should be right.. i have my calculus exam tommorow.

hmmm, you sure? when I graph it, it approaches 0 not .75


also what do i do for the limit as it approaches negative infinity?

 

[simms]

 

[Syringer]

Originally posted by: simms
divide everything by x. ... crap this is gonna be tough. pretend the root is there..
9x+2/4x+3

= lim x->8 (9x/x + 2/x) / (4x/x + 3/x)
= lim x->8 (9 + 2/x) / (4 + 3/x)
= (9+ 0 [which is zero, 2/infinity is zero]) / (4 + 0)

= rt (9) / 4
= 3/4

The limit is 3/4.

Answer according to TI-89 it's actually zero. I don't think you can do what you did with the sqrt function up top..

 

[wildwarren]

scratch what I said....

Square the top and bottom, then divide by the x^2 you get in the bottom giving you two zero's on top?

 

[Syringer]

Hmm, actually L'Hospital's rule won't work since the numerator is a sqrt, so the derivatives would go on forever.

But you can just tell it's going to zero since the denominator is increasing much faster than the numerator is..

 

[Syringer]

Originally posted by: wildwarren
scratch what I said....

Square the top and bottom, then divide by the x^2 you get in the bottom giving you two zero's on top?

You can't square the top and bottom..because for example 3/4 doesn't equal 9/16..

The only way to get rid of the radical is to multiply the top and bottom by sqrt(9x+2), which doesn't do anything.

 

[skim milk]

what if you broke up the two pieces?

do the limit of numerator and denominator separately

the limit for the top is 0
the limit for the bottom is infinity

which is saying limit is equal to 0/infinity = 0?

 

[alphatarget1]

0

lim h>0
9(x+h)+2-9x-2/4(x+h)+3-4x-3

i think....

EDIT: ok nvm, that's the derivative.

they're both approaching infinity, so infinity/infinity=0 (look up limit laws)

 

[Syringer]

Originally posted by: fritolays
what if you broke up the two pieces?

do the limit of numerator and denominator separately

the limit for the top is 0
the limit for the bottom is infinity

which is saying limit is equal to 0/infinity = 0?

The limit for the top is infinity too..

 

[simms]

got it. it is 0. The limit goes to 0.. the numerator is constant, while the denominator goes to (8).

 

[skim milk]

Originally posted by: simms
got it. it is 0. The limit goes to 0.. the numerator is constant, while the denominator goes to (8).

yeah, i got zero from just graphing... but how can it be worked out without a calculator?

 

[Syringer]

You can just tell it's going to zero since the denominator is increasing faster than the numerator is..

 

[skim milk]

Originally posted by: Syringer
You can just tell it's going to zero since the denominator is increasing faster than the numerator is..

so on the next test or whatever, no works needs to be shown?


how about the negative infinity, what limit does that approach? it looks like it "DNE"

can someone confirm? thanks...

 

[simms]

Originally posted by: Syringer
You can just tell it's going to zero since the denominator is increasing faster than the numerator is..

that's what i did too. you can multiply by conjugate base, reduce, and you'll get a constant C:

lim welajfdsj C/ rt x + C

so as x goes to 8, the denominator gets bigger -> 0

 

[Syringer]

Originally posted by: fritolays

Originally posted by: Syringer
You can just tell it's going to zero since the denominator is increasing faster than the numerator is..

so on the next test or whatever, no works needs to be shown?


how about the negative infinity, what limit does that approach? it looks like it "DNE"

can someone confirm? thanks...

Well since there's a positive x term in the sqrt, anything negative would make it undefined.