## nickymarden 3 years ago A question from my test today...

1. nickymarden

2. nickymarden

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

3. eigenschmeigen

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

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

5. eigenschmeigen

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

6. nickymarden

yup. :)

7. eigenschmeigen

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

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

9. eigenschmeigen

hmm

10. eigenschmeigen

call in the cavalry! @FoolForMath

11. FoolForMath

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

12. eigenschmeigen

aw ok

13. eigenschmeigen

@TuringTest

14. TuringTest

whoa what happened? let em read this

15. nickymarden

haha

16. TuringTest

the limit as x goes to 1 ?

17. zepp

Apply the squeeze theorem to the question :)

18. TuringTest

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

19. nickymarden

$\lim_{x \rightarrow 1}$

20. experimentX

looks like 2

21. TuringTest

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

22. eigenschmeigen

i factorised wrong i think

23. TuringTest

oh I trusted you! ok I'll check

24. eigenschmeigen

derp

25. eigenschmeigen

sorry guys

26. experimentX

-(x-1)(x-3)

27. eigenschmeigen

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

28. SmoothMath

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

in fact ima go sleep night!

30. SmoothMath

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

31. TuringTest

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

32. experimentX

well that was delicious.

33. nickymarden

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

34. SmoothMath

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

35. nickymarden

i knoow.

36. experimentX

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

37. TuringTest

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

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

39. SmoothMath

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

40. experimentX

real mathematicians do that ..!!!

41. SmoothMath

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

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

43. experimentX

you are indeed a real one!!!

44. TuringTest

agreed :)

45. SmoothMath

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

haha you should be "SexyMath" then

47. nickymarden

omg

48. experimentX

LOL..

49. nickymarden

thanks :)

50. SmoothMath

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