Yarrrgh I'm helping this guy with calc 1 and apparently I suck now

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    Calc 1
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Homework Help Overview

The discussion revolves around limits in calculus, specifically focusing on evaluating limits involving cube roots and square roots. The original poster expresses difficulty with these concepts and seeks assistance in finding values that satisfy given limit conditions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore various methods for simplifying expressions involving radicals, such as multiplying out terms and considering binomial expansions. There is also discussion about the possibility of finding functions that meet specific limit criteria despite individual limits not existing.

Discussion Status

Some participants have offered suggestions for tackling the problems, while others express confusion and seek further clarification. The conversation indicates a mix of attempts to resolve the limit problems and exploration of underlying concepts.

Contextual Notes

Participants note the constraint of not using derivatives or l'Hôpital's Rule, which shapes their approaches to the limit problems. There is also a mention of the original poster's skepticism regarding the existence of certain functions.

schattenjaeger
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Limits:( If it were things like derivatives and integrals I'd be good to go but nooooo, stupid limits

lim(x-->0) [(1+ax)^(1/3)-1]/x=1

find a such that this is true. Guessing on my calculator tells me its 3 but I don't know how to do it analytically. My first guess is clear the radical, but I don't know how to clear a cube root without making it worse, if you can

lim(x-->2)[sqrt(6-x)-2]/[sqrt(3-x)-1]=?

again, calculator tells me 1/2, but I can't. Trying to clear either numerator or denominator of its radicals gets either the top or bottom to be 2-x and the other side to be a bigger mess which I can decipher.

and finally, pick two functions f(x) and g(x) such that lim(x-->0)[f(x)+g(x)] exists, but f(x) and g(x)'s limits as x-->0 DON'T exist

I really don't think that's possible and wouldn't know where to start finding those functions if it weren't
 
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Oh, and no derivatives yet, so no l'Hopital's Rule
 
Expand the cube root bit as a binomial series.
 
Actually I just figured that one out

First step is multiply out that x, then subtract the 1, cube both sides, and everything falls into place from there

the second one has me MEGAstumped though
 
I would get rid of the square root term on the top...

I'll let you have a go at doing this.
 
Getting rid of either square root doesn't seem to help. Like I said you get 2-x on the side you do it on and some ridiculous string of stuff that still ends up as 0.
 
Like you saud, if you try to get rid if the square root above, you get 2-x in the numerator. If you try to get rid of the square root below, you also get 2-x, but in the denominator. That should give you an idea on what to do. What if you got 2-x on both sides?
 
The third one is trivial. Take some f with no limit as x->0, say sin(1/x), and let g = -f :-p
 
I KNEW it

always happens, problems where my first instinct is "hey that's not possible" usually have really simple solutions.
 

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