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A question from my test today...

Mathematics
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can't anyone do it? Where's @TuringTest
can we factorise and cancel to get rid of the pesky denominator? :\[\frac{-x(x-1)(x+3)}{x-1} \le g(x) \le \frac{(x+1)(x-1)}{x-1}\]

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Other answers:

i guess, i didnt have time to do it, Last question of the test... ;/
so: \[-x(x+3) \le g(x) \le (x+1)\] and you want the limit as x tends to 1 ?
yup. :)
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
it isn't, there's only one value. ;/ But thank you :))
hmm
call in the cavalry! @FoolForMath
My mind is officially fried for the day :( sorry.
aw ok
whoa what happened? let em read this
haha
the limit as x goes to 1 ?
Apply the squeeze theorem to the question :)
what limit is this, I don't understand the notation
\[\lim_{x \rightarrow 1}\]
looks like 2
I hope smoothmath's cookin up something good 'cuz I see what Eigen saw... how do you figure X ?
i factorised wrong i think
oh I trusted you! ok I'll check
derp
sorry guys
-(x-1)(x-3)
yah. sorry im exhausted, just finished ~ 4hrs physics work
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.
in fact ima go sleep night!
Haha don't worry, Turing. I cooked up something delicious.
I can see that! that's what I get for not double-checking other peoples algebra...
well that was delicious.
good night, and thank you :P We can't use L'hopital's rule..he said he wouldn't consider the answer ;/
Ugh. Sucks. I hate that he disallowed it because it works so well...
i knoow.
factorize and cancel ... that's quite easier that L'hospital's rule.
but it does factor to\[{-x(x-1)(x-3)\over x-1}\le g(x)\le x+1\]\[-x(x-3)\le g(x)\le x+1\]so yeah, we can just plug in like experimentX says
well ... still, I really liked that squeezing theorem.
Haha that's a matter of opinion, experiment. I much prefer taking derivatives of simple polynomials to factoring them.
real mathematicians do that ..!!!
But yeah, to solve by factoring, turing's method above is correct. Eigen's original misfactoring is what gave the error we got.
Sorry. Real mathematicians do what? Are you saying I'm a real mathematician? Or not one? I'm confused.
you are indeed a real one!!!
agreed :)
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.
haha you should be "SexyMath" then
omg
LOL..
thanks :)
lols. That might give the wrong impression, but I like it.

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