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anonymous
 4 years ago
A question from my test today...
anonymous
 4 years ago
A question from my test today...

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0call in the cavalry! @FoolForMath

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

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1the limit as x goes to 1 ?

zepp
 4 years ago
Best ResponseYou've already chosen the best response.0Apply the squeeze theorem to the question :)

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i factorised wrong i think

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in fact ima go sleep night!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha don't worry, Turing. I cooked up something delicious.

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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0well that was delicious.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ugh. Sucks. I hate that he disallowed it because it works so well...

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

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1but 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha that's a matter of opinion, experiment. I much prefer taking derivatives of simple polynomials to factoring them.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0real mathematicians do that ..!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But yeah, to solve by factoring, turing's method above is correct. Eigen's original misfactoring is what gave the error we got.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry. Real mathematicians do what? Are you saying I'm a real mathematician? Or not one? I'm confused.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0you are indeed a real one!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha 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.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1haha you should be "SexyMath" then

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lols. That might give the wrong impression, but I like it.
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