Question

A question from my test today...

Answer 1

can't anyone do it? Where's @TuringTest

Answer 2

can we factorise and cancel to get rid of the pesky denominator? :\[\frac{-x(x-1)(x+3)}{x-1} \le g(x) \le \frac{(x+1)(x-1)}{x-1}\]

Answer 3

i guess, i didnt have time to do it, Last question of the test... ;/

Answer 4

so:

\[-x(x+3) \le g(x) \le (x+1)\]

and you want the limit as x tends to 1 ?

Answer 5

yup. :)

Answer 6

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

Answer 7

it isn't, there's only one value. ;/ But thank you :))

Answer 8

hmm

Answer 9

call in the cavalry! @FoolForMath

Answer 10

My mind is officially fried for the day :( sorry.

Answer 11

aw ok

Answer 12

@TuringTest

Answer 13

whoa what happened?
let em read this

Answer 14

haha

Answer 15

the limit as x goes to 1 ?

Answer 16

Apply the squeeze theorem to the question :)

Answer 17

what limit is this, I don't understand the notation

Answer 18

\[\lim_{x \rightarrow 1}\]

Answer 19

looks like 2

Answer 20

I hope smoothmath's cookin up something good 'cuz I see what Eigen saw...
how do you figure X ?

Answer 21

i factorised wrong i think

Answer 22

oh I trusted you!
ok I'll check

Answer 23

derp

Answer 24

sorry guys

Answer 25

-(x-1)(x-3)

Answer 26

yah. sorry im exhausted, just finished ~ 4hrs physics work

Answer 27

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.

Answer 28

in fact ima go sleep night!

Answer 29

Haha don't worry, Turing. I cooked up something delicious.

Answer 30

I can see that!
that's what I get for not double-checking other peoples algebra...

Answer 31

well that was delicious.

Answer 32

good night, and thank you :P We can't use L'hopital's rule..he said he wouldn't consider the answer ;/

Answer 33

Ugh. Sucks. I hate that he disallowed it because it works so well...

Answer 34

i knoow.

Answer 35

factorize and cancel ... that's quite easier that L'hospital's rule.

Answer 36

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

Answer 37

well ... still, I really liked that squeezing theorem.

Answer 38

Haha that's a matter of opinion, experiment. I much prefer taking derivatives of simple polynomials to factoring them.

Answer 39

real mathematicians do that ..!!!

Answer 40

But yeah, to solve by factoring,
turing's method above is correct. Eigen's original misfactoring is what gave the error we got.

Answer 41

Sorry. Real mathematicians do what? Are you saying I'm a real mathematician? Or not one? I'm confused.

Answer 42

you are indeed a real one!!!

Answer 43

agreed :)

Answer 44

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.

Answer 45

haha you should be "SexyMath" then

Answer 46

omg

Answer 47

LOL..

Answer 48

thanks :)

Answer 49

lols. That might give the wrong impression, but I like it.