## nickymarden Group Title A question from my test today... 2 years ago 2 years ago

1. nickymarden Group Title

2. nickymarden Group Title

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

3. eigenschmeigen Group Title

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

4. nickymarden Group Title

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

5. eigenschmeigen Group Title

so: $-x(x+3) \le g(x) \le (x+1)$ and you want the limit as x tends to 1 ?

6. nickymarden Group Title

yup. :)

7. eigenschmeigen Group Title

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

8. nickymarden Group Title

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

9. eigenschmeigen Group Title

hmm

10. eigenschmeigen Group Title

call in the cavalry! @FoolForMath

11. FoolForMath Group Title

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

12. eigenschmeigen Group Title

aw ok

13. eigenschmeigen Group Title

@TuringTest

14. TuringTest Group Title

whoa what happened? let em read this

15. nickymarden Group Title

haha

16. TuringTest Group Title

the limit as x goes to 1 ?

17. zepp Group Title

Apply the squeeze theorem to the question :)

18. TuringTest Group Title

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

19. nickymarden Group Title

$\lim_{x \rightarrow 1}$

20. experimentX Group Title

looks like 2

21. TuringTest Group Title

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

22. eigenschmeigen Group Title

i factorised wrong i think

23. TuringTest Group Title

oh I trusted you! ok I'll check

24. eigenschmeigen Group Title

derp

25. eigenschmeigen Group Title

sorry guys

26. experimentX Group Title

-(x-1)(x-3)

27. eigenschmeigen Group Title

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

28. SmoothMath Group Title

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.

29. eigenschmeigen Group Title

in fact ima go sleep night!

30. SmoothMath Group Title

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

31. TuringTest Group Title

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

32. experimentX Group Title

well that was delicious.

33. nickymarden Group Title

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

34. SmoothMath Group Title

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

35. nickymarden Group Title

i knoow.

36. experimentX Group Title

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

37. TuringTest Group Title

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

38. experimentX Group Title

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

39. SmoothMath Group Title

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

40. experimentX Group Title

real mathematicians do that ..!!!

41. SmoothMath Group Title

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

42. SmoothMath Group Title

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

43. experimentX Group Title

you are indeed a real one!!!

44. TuringTest Group Title

agreed :)

45. SmoothMath Group Title

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.

46. TuringTest Group Title

haha you should be "SexyMath" then

47. nickymarden Group Title

omg

48. experimentX Group Title

LOL..

49. nickymarden Group Title

thanks :)

50. SmoothMath Group Title

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