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 Post subject: limits also on with trigPosted: Sat, 2 Oct 2010 09:26:13 UTC
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Joined: Sat, 2 Oct 2010 09:19:23 UTC
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lim x->0 [sqrt(x+9)-3] /x goes to 1/6 and i have no clue how

lim x->0 (x-sinx)/x^3 also goes to 1/6 and i have no clue how really need

help im taking enriched calculus 1

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 Post subject: Posted: Sat, 2 Oct 2010 09:29:36 UTC
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Joined: Sat, 2 Oct 2010 09:19:23 UTC
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Location: kamloops
well i mean i think ive come close but cant quite get it to make sense

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 Post subject: Re: limits also on with trigPosted: Sat, 2 Oct 2010 10:45:02 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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brodie6 wrote:
lim x->0 [sqrt(x+9)-3] /x goes to 1/6 and i have no clue how

lim x->0 (x-sinx)/x^3 also goes to 1/6 and i have no clue how really need

help im taking enriched calculus 1

Hint for the first question:

Hint for the seond question: Apply (one of the many) definitions of sin(x).

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 Post subject: Posted: Sat, 2 Oct 2010 18:27:51 UTC
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Joined: Sat, 2 Oct 2010 09:19:23 UTC
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for the first one that was deffinatly my first move and it all makes sense now it was just late and i was missing that the x s cross once you get 9-9 = 0 and the bottom goes to 6 when 3+3 i am thinking of this correctly right ? , for the trig one i dont understand any of the notation your using ive tried setting sinx to sqrt(1-cox^2) which i think is close since i think it will have to be a 1+cosx twice on the bottom or ive tried the conjugate top and bottom by x+sinx to deal with sinx^2 instead but i just run into zeros this way as well and just feels like im going in circles once i get x- sqrt(1-cosx^2) am i allowed to go conjugate sqrt(1+cosx^2) then use the ratio of the x/x^3 to get 1/3 then go 3(2) is this the right track

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 Post subject: Posted: Sun, 3 Oct 2010 05:44:13 UTC
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Which rigorous definition of sin(x) are you using?

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 Post subject: Posted: Sun, 3 Oct 2010 06:08:39 UTC
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i cant tell if your being sarcastic about the rigorous deffinition but ive been trying sqrt(1-cox^2) and sqrt(1-cos2x/2) is this not deep enough i must be a fool cuz it has me losing sleep and i can do mostly every other limit problem in my book or i keep doing the same mistake over and over

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 Post subject: Posted: Sun, 3 Oct 2010 06:25:41 UTC
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brodie6 wrote:
i cant tell if your being sarcastic about the rigorous deffinition but ive been trying sqrt(1-cox^2) and sqrt(1-cos2x/2) is this not deep enough i must be a fool cuz it has me losing sleep and i can do mostly every other limit problem in my book or i keep doing the same mistake over and over

One definition of is for all . See what you can get from here.

(There are other equivalent definitions of the sine, such as using Euler's sine product or uniqueness to solutions of differential equations. But to say "sin(x) is the y-coordinates of ... when the angle is x", without a definition of angle that doesn't make the definition circular, is not a definition of sin(x).)

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 Post subject: Posted: Mon, 4 Oct 2010 05:09:24 UTC
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Joined: Fri, 27 Jul 2007 10:17:26 UTC
Posts: 278
Location: Chandler, AZ, USA
Outermeasure's just trying to psych you out - you won't see that definition of sine for a long time, since the problem you are doing is first semester Calculus.

Just split up into two fractions:

Check your answer - you should be able to get each of the two resulting limits to be zero, especially since you should just have learned a limit theorem involving . Shouldn't be .

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 Post subject: Posted: Mon, 4 Oct 2010 05:23:16 UTC
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alstat wrote:
Outermeasure's just trying to psych you out - you won't see that definition of sine for a long time, since the problem you are doing is first semester Calculus.

Just split up into two fractions:

Check your answer - you should be able to get each of the two resulting limits to be zero, especially since you should just have learned a limit theorem involving . Shouldn't be .

No! The limit is indeed 1/6. The poles parts cancel, but you need to consider the finite part of , namely the order x^2 part of .

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 Post subject: Posted: Mon, 4 Oct 2010 12:11:31 UTC
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For the second one, use L'HÃ´pital's rule three times in a row.

Outermeasure's suggestion was to write sin x as the infinite series , a method which works perfectly well, but is probably not covered in your course.

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 Post subject: Posted: Tue, 5 Oct 2010 01:45:54 UTC
 S.O.S. Oldtimer

Joined: Fri, 27 Jul 2007 10:17:26 UTC
Posts: 278
Location: Chandler, AZ, USA
I was wrong! My apologies - never try this on a lack of sleep!
My question is, is L'Hopital's rule available in Calculus 1?
It usually is only about a chapter ahead of the infinte series that would also solve it. Usually both L'Hopital & series are in 2nd semester Calc.

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 Post subject: Posted: Tue, 5 Oct 2010 01:57:36 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
alstat wrote:
My question is, is L'Hopital's rule available in Calculus 1?

Depends on who's teaching it.

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