anonymous 4 years ago help with AP calc hw plz!

1. Hero

Simply posting your actual question would be great

2. anonymous

im going to take a pic :3

3. anonymous

4. anonymous

the limit is x->-1/2+

5. Hero

You've got like 5 questions circled. Which one is it?

6. anonymous

#38

7. campbell_st

perhaps factorising the numerator and denominator will allow you to cancel a common factor

8. anonymous

kk

9. anonymous

limit as x goes to ???

10. Hero

-1/2

11. campbell_st

there is a common factor that is evident when you factorise numerator and denominator

12. anonymous

then replace $$x$$ by $$-\frac{1}{2}$$ if you get a number, that is your answer any rational function is continuous on its domain so the limit and the value are the same

13. Hero

I think @campbell_st is right. You should try factoring the numerator and denominator first before plugging in x = -1/2

14. anonymous

if you get $$\frac{2}{0}$$ then you know you can factor both top and bottom as $$(2x+1)(\text{something)}$$

15. anonymous

i wouldn't do that, i would plug in first

16. campbell_st

the other thing about eliminating a common factor is that it makes the substitution calculation easier

17. anonymous

if you get something other than $$\frac{0}{0}$$ then that is the answer if you do get $$\frac{0}{0}$$ then you know how to factor

18. Hero

If you plug in first, you're definitely going to get a denominator you won't like

19. anonymous

@campbell_st i suck at factoring, so i like to know if i can factor first if i know i have a zero of a polynomial, i know already how to factor it that is why i would check first and not waste my time factoring, in case for example it does not factor over the integers

20. anonymous

so plugin -1/2 im kinda confused

21. Hero

Why waste your time trying to evaluate before factoring? Seems silly

22. campbell_st

lol... people get lost with squaring and substituting negative fractions... its really what works best you each of us

23. anonymous

what i am trying to say is that if i replace $$x$$ by $$-\frac{1}{2}$$ i know it must factor as $$(2x+1)\times (\text{whatever})$$

24. Zarkon

rational functions are continuous on their domain..if you plug in the value and you get a real number then that is the answer to the limit problem

25. anonymous

because it tells me exactly how to factor otherwise i am doing the dance of what two numbers blah blah blah

26. campbell_st

here is the factorised version I hope... $\lim_{x \rightarrow -\frac{1}{2}} \frac{(2x + 1)(3x -1)}{(2x + 1)(2x - 3)}$

27. anonymous

kk

28. Hero

@satellite lost me. You're telling me that plugging in x = -1/2 tells you which two numbers you use to factor it?

29. anonymous

@hero both @zarkon and i answered the question "why evaluate first" i think

30. anonymous

yes, by the "factor theorem" if $$r$$ is a root of a polynomial $$p(x)$$ then we know $$p(x)=(x-r)q(x)$$

31. Hero

No, @zarkon said something I already knew. The problem is, if you plug in x = -1/2 in this case, you clearly get a zero denominator

32. Agent47

This limit is begging for L'Hopital's Rule.....

33. anonymous

it is not some miracle that each time you get $$\frac{0}{0}$$ you can factor and cancel. it is precisely because of the factor theorem. you HAVE to be able to factor out the zeros

34. campbell_st

it doesn't need L'Hopital ... just factor to find it.... don't over think it

35. anonymous

im kinda confuzzled a bit :3

36. Agent47

i can do lhopital in my head on this... but i guess it depends

37. Hero

That's because there's too many people trying to give their version of how to solve it.

38. anonymous

what i am trying to say is this: you want the limit of a rational function as x approaches some number first thing you should do is check what you get when you evaluate the function at that number if you get a number back, that is your answer and you are done

39. campbell_st

looking at the type of questions you need to factorise to find the limit..

40. Agent47

abanna, just listen to satellite, he's always right lol

41. Hero

I object to that statement^

42. anonymous

if you get $$\frac{a}{0}$$ where $$a$$ is some non zero number, then there is no limit

43. campbell_st

@abannavong have a look above at my factorised version.. eliminate the common factor then substitute for your limit.

44. anonymous

and if you get $$\frac{0}{0}$$ then you know to factor and cancel AND you know how to factor

45. Hero

I see what you're saying now, @satellite. I usually do that part in my head first, but not on paper.

46. anonymous

oh ok

47. anonymous

because for example, in this case we see that $$-\frac{1}{2}$$ gives 0 that tells us that it MUST factor as $$(2x+1)(\text{whatever})$$ because $$2x+1$$ has a zero at $$-\frac{1}{2}$$

48. campbell_st

@abannavong what level of maths is this...

49. anonymous

and your job is then only to find the other factor, as @campbell_st wrote above

50. Agent47

AB calculus, aka cal 1

51. anonymous

AP Calc :3

52. campbell_st

ok... so still high school... just I'm in australia...

53. anonymous

yeah in grade 12

54. campbell_st

thanks... I think you get 5/8 as the limit... good luck

55. anonymous

thank you @campbell_st