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nickymardenBest ResponseYou've already chosen the best response.0
can't anyone do it? Where's @TuringTest
 2 years ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
can we factorise and cancel to get rid of the pesky denominator? :\[\frac{x(x1)(x+3)}{x1} \le g(x) \le \frac{(x+1)(x1)}{x1}\]
 2 years ago

nickymardenBest ResponseYou've already chosen the best response.0
i guess, i didnt have time to do it, Last question of the test... ;/
 2 years ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
so: \[x(x+3) \le g(x) \le (x+1)\] and you want the limit as x tends to 1 ?
 2 years ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
well evaluating at x = 1 we have \[4 \le g(x) \le 2\] im not sure whether thats what you want or not, i haven't really seen this kinda Q before
 2 years ago

nickymardenBest ResponseYou've already chosen the best response.0
it isn't, there's only one value. ;/ But thank you :))
 2 years ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
call in the cavalry! @FoolForMath
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
My mind is officially fried for the day :( sorry.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
whoa what happened? let em read this
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
the limit as x goes to 1 ?
 2 years ago

zeppBest ResponseYou've already chosen the best response.0
Apply the squeeze theorem to the question :)
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
what limit is this, I don't understand the notation
 2 years ago

nickymardenBest ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow 1}\]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
I hope smoothmath's cookin up something good 'cuz I see what Eigen saw... how do you figure X ?
 2 years ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
i factorised wrong i think
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
oh I trusted you! ok I'll check
 2 years ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
yah. sorry im exhausted, just finished ~ 4hrs physics work
 2 years ago

SmoothMathBest ResponseYou've already chosen the best response.3
I like l'hopitals rule for both of those equations. To use L'hopital's rule, we have to make sure they satisfy the condition that the functions at that point are of the indeterminate form 0/0 or inf/inf. These both look like 0/0, so we're good to go. Now, since we've satisfied that condition, L'hopital's rule says that the limit of each function will be the same as if we take the derivative of the top and bottom. So basically, if our function is f(x)/g(x), the rule says that the limit will be the same as f'(x)/g'(x), as long as that limit exists. What that gives us is: (3x^2 + 8x  3)/1 <= g(x) <= 2x/1 Evaluating at x=1 gives: (3 + 8 3)/1 <= g(x) <= 2/1 2 <=g(x) <=2 So, by squeeze theorem, lim g(x) is 2.
 2 years ago

eigenschmeigenBest ResponseYou've already chosen the best response.0
in fact ima go sleep night!
 2 years ago

SmoothMathBest ResponseYou've already chosen the best response.3
Haha don't worry, Turing. I cooked up something delicious.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
I can see that! that's what I get for not doublechecking other peoples algebra...
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
well that was delicious.
 2 years ago

nickymardenBest ResponseYou've already chosen the best response.0
good night, and thank you :P We can't use L'hopital's rule..he said he wouldn't consider the answer ;/
 2 years ago

SmoothMathBest ResponseYou've already chosen the best response.3
Ugh. Sucks. I hate that he disallowed it because it works so well...
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
factorize and cancel ... that's quite easier that L'hospital's rule.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
but it does factor to\[{x(x1)(x3)\over x1}\le g(x)\le x+1\]\[x(x3)\le g(x)\le x+1\]so yeah, we can just plug in like experimentX says
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
well ... still, I really liked that squeezing theorem.
 2 years ago

SmoothMathBest ResponseYou've already chosen the best response.3
Haha that's a matter of opinion, experiment. I much prefer taking derivatives of simple polynomials to factoring them.
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
real mathematicians do that ..!!!
 2 years ago

SmoothMathBest ResponseYou've already chosen the best response.3
But yeah, to solve by factoring, turing's method above is correct. Eigen's original misfactoring is what gave the error we got.
 2 years ago

SmoothMathBest ResponseYou've already chosen the best response.3
Sorry. Real mathematicians do what? Are you saying I'm a real mathematician? Or not one? I'm confused.
 2 years ago

experimentXBest ResponseYou've already chosen the best response.0
you are indeed a real one!!!
 2 years ago

SmoothMathBest ResponseYou've already chosen the best response.3
Haha okay. Cool. Both methods are great, and I actually think factoring is easier to teach. I just find L'hopital's easier to execute. And sexier.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.1
haha you should be "SexyMath" then
 2 years ago

SmoothMathBest ResponseYou've already chosen the best response.3
lols. That might give the wrong impression, but I like it.
 2 years ago
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