# I need help with a couple calculus questions.

Discussion in 'The Lounge' started by colbystephens, Feb 27, 2006.

1. ### colbystephens1 ton statusGMOTM Winner

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not real tough stuff here. anyway, i'm trying to find the limit of (t[sup]2[/sup]-4)/(t[sup]3[/sup]-8) as t tends to 2. i can't figure out the generic form of a[sup]3[/sup]-b[sup]3[/sup] so i can determine how things cancel out. by generic form i'm referring to, as an example, (a[sup]2[/sup] - b[sup]2[/sup]= (a-b)(a+b).

also, i'm looking for the limit of (sqrt(x[sup]2[/sup]+x+1)-sqrt(x[sup]2[/sup]-1)) as x tends to positive infinity. when i calculate it, i get + or - 1/2. can someone verify that result for me?

thanks!

2. ### colbystephens1 ton statusGMOTM Winner

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hmmm. looks like my html code doesn't work. i'll try and fix that. for now, where it says [sup]2[/sup] i'm referring to "squared" or "cubed" depending on the number between the code. thanks!

3. ### colbystephens1 ton statusGMOTM Winner

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ok. i don't know how to format it. sorry.

4. ### kyser_soze1/2 ton status

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use this x^2, or x^3

5. ### colbystephens1 ton statusGMOTM Winner

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ok - got the first one figured out. can someone still verify my other question tho?

6. ### jekquistk5Weld nekidPremium MemberGMOTM Winner

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I'm still confused on your html code or else I could help

ok give me a couple minutes but from looking at if a little harder I got

(sqrt(x[sup]2[/sup]+x+1)-sqrt(x[sup]2[/sup]-1)) =

Lim x->+ infinity sqrt((x^2)+x+1)-sqrt((x^2)-1) and you can use the limit laws to break that up to lim x->+ inifinity sqrt((x^2)+x+1) - Lim x-> sqrt((x^2)-1)

ahh my brain is fried right now and its been 2 yrs since calc

Ok you probably dont want to do that because that puts it into an indeterminate form. Good luck sorry I'm no help.. but atleast I tried

7. ### colbystephens1 ton statusGMOTM Winner

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thanks for the try. i found out that my answer is not + or -, it's just + because i'm going for positive infinity.

8. ### Jagged1 ton status

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Second answer... approaches (+)infinity.

x^2+x+1 > x-1

9. ### colbystephens1 ton statusGMOTM Winner

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i'm pretty sure that's not correct. as x tends to infinity, it should tend to +1/2. but, if you care to explain, i'd be open to objections.

10. ### Jagged1 ton status

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Power rule

x^n > x^m => lim[x->inf] ( x^n - x^m ) = inf (sqare root of x^2 is x, and x is x^1/2)

Also:
lim[x->inf] x^n = ( lim[x->inf] x ) ^ n is true IIRC

edit:
Im a dumb****... didnt see the second part is x^2... thought it was x

oops... might be 1/2 then

11. ### Jagged1 ton status

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Hrm... what's your proof for 1/2?

I think I can prove infinity with the squeeze theorem.

I just broke out my calculus book

12. ### colbystephens1 ton statusGMOTM Winner

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after multiplying by the conjugate, and cancelling, you end up with x/(sqrt(x^2 + x + 1) + sqrt(x^2 - 1)), then pull out an x^2 from the denominator so that you have x/(sqrt[x^2(1 + (x/x^2) + (1/x^2)] + sqrt[x^2(1 - (1/x^2))]. Then, one can take the root of x^2 so you end up with x/[(x)(sqrt(1 + (x/x^2)+ (1/x^2))+sqrt(1 - (1/x^2)]. Then you separate it so that x/x=1 and you are left with 1/[(sqrt(1 + (x/x^2)+ (1/x^2))+sqrt(1 - (1/x^2)]. Then, factoring infinity in for x, everything with an x denominator effectively goes to 0 and you're left with 1/(sqrt(1)+sqrt(1)) which equals 1/2. Notice that the x in red above should have been pulled out as an absolute value, but since we are only dealing with positive infinity, only the positve x need be considered.

13. ### az-k51/2 ton status

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I got infinity doing it in my head. What calc is this? I II or III (so I can kinda know what rules you are allowed to use)

Even if they approach infinity at differing rates, it is going from the positive side so that negates a bunch. The intermidiate found using the limit laws allows for an 'infinity - infinity' situation wich you can L'Hptel (sp = bad) down to infinity - infinty again giving the limit is zero. Do it again and you have infinty. (I am still to lazy to find a pencil and paper so this is just from memory and brain work)

14. ### colbystephens1 ton statusGMOTM Winner

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effectively calc 1

15. ### Jagged1 ton status

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L'Hopital!!!!! I forgot that rule.

You can prove it with L'Hopital easily

(the square root really has no affect on the outcome of this problem... just a small nusance for using the proper limit laws)

Get rid of the x^2
x+1 > -1
Infinity

16. ### colbystephens1 ton statusGMOTM Winner

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not familiar with any of that new age hippity-hoppity. care to explain?

17. ### Jagged1 ton status

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Erm.. forgot the whole rule

2x+1 > 1
Infinity

Anywho, I'm gonna take a picture for you

18. ### thezentree3/4 ton status

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That actually made smoke come out of my ears. Seriously, I'm worried it's gonna set the smoke detector off.

19. ### Jagged1 ton status

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Note: You can use L'Hostpal recursively....

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