## abannavong Group Title help with AP calc hw plz! one year ago one year ago

1. Hero Group Title

Simply posting your actual question would be great

2. abannavong Group Title

im going to take a pic :3

3. abannavong Group Title

4. abannavong Group Title

the limit is x->-1/2+

5. Hero Group Title

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

6. abannavong Group Title

#38

7. campbell_st Group Title

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

8. abannavong Group Title

kk

9. satellite73 Group Title

limit as x goes to ???

10. Hero Group Title

-1/2

11. campbell_st Group Title

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

12. satellite73 Group Title

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 Group Title

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

14. satellite73 Group Title

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

15. satellite73 Group Title

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

16. campbell_st Group Title

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

17. satellite73 Group Title

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 Group Title

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

19. satellite73 Group Title

@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. abannavong Group Title

so plugin -1/2 im kinda confused

21. Hero Group Title

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

22. campbell_st Group Title

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

23. satellite73 Group Title

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 Group Title

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. satellite73 Group Title

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

26. campbell_st Group Title

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

27. abannavong Group Title

kk

28. Hero Group Title

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

29. satellite73 Group Title

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

30. satellite73 Group Title

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 Group Title

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 Group Title

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

33. satellite73 Group Title

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 Group Title

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

35. abannavong Group Title

im kinda confuzzled a bit :3

36. Agent47 Group Title

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

37. Hero Group Title

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

38. satellite73 Group Title

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 Group Title

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

40. Agent47 Group Title

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

41. Hero Group Title

I object to that statement^

42. satellite73 Group Title

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

43. campbell_st Group Title

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

44. satellite73 Group Title

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

45. Hero Group Title

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

46. abannavong Group Title

oh ok

47. satellite73 Group Title

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 Group Title

@abannavong what level of maths is this...

49. satellite73 Group Title

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

50. Agent47 Group Title

AB calculus, aka cal 1

51. abannavong Group Title

AP Calc :3

52. campbell_st Group Title

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

53. abannavong Group Title