anonymous
  • anonymous
help with AP calc hw plz!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Hero
  • Hero
Simply posting your actual question would be great
anonymous
  • anonymous
im going to take a pic :3
anonymous
  • anonymous

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anonymous
  • anonymous
the limit is x->-1/2+
Hero
  • Hero
You've got like 5 questions circled. Which one is it?
anonymous
  • anonymous
#38
campbell_st
  • campbell_st
perhaps factorising the numerator and denominator will allow you to cancel a common factor
anonymous
  • anonymous
kk
anonymous
  • anonymous
limit as x goes to ???
Hero
  • Hero
-1/2
campbell_st
  • campbell_st
there is a common factor that is evident when you factorise numerator and denominator
anonymous
  • 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
Hero
  • Hero
I think @campbell_st is right. You should try factoring the numerator and denominator first before plugging in x = -1/2
anonymous
  • anonymous
if you get \(\frac{2}{0}\) then you know you can factor both top and bottom as \((2x+1)(\text{something)}\)
anonymous
  • anonymous
i wouldn't do that, i would plug in first
campbell_st
  • campbell_st
the other thing about eliminating a common factor is that it makes the substitution calculation easier
anonymous
  • 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
Hero
  • Hero
If you plug in first, you're definitely going to get a denominator you won't like
anonymous
  • 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
anonymous
  • anonymous
so plugin -1/2 im kinda confused
Hero
  • Hero
Why waste your time trying to evaluate before factoring? Seems silly
campbell_st
  • campbell_st
lol... people get lost with squaring and substituting negative fractions... its really what works best you each of us
anonymous
  • anonymous
what i am trying to say is that if i replace \(x\) by \(-\frac{1}{2}\) i know it must factor as \((2x+1)\times (\text{whatever})\)
Zarkon
  • 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
anonymous
  • anonymous
because it tells me exactly how to factor otherwise i am doing the dance of what two numbers blah blah blah
campbell_st
  • campbell_st
here is the factorised version I hope... \[\lim_{x \rightarrow -\frac{1}{2}} \frac{(2x + 1)(3x -1)}{(2x + 1)(2x - 3)}\]
anonymous
  • anonymous
kk
Hero
  • Hero
@satellite lost me. You're telling me that plugging in x = -1/2 tells you which two numbers you use to factor it?
anonymous
  • anonymous
@hero both @zarkon and i answered the question "why evaluate first" i think
anonymous
  • anonymous
yes, by the "factor theorem" if \(r\) is a root of a polynomial \(p(x)\) then we know \(p(x)=(x-r)q(x)\)
Hero
  • 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
Agent47
  • Agent47
This limit is begging for L'Hopital's Rule.....
anonymous
  • 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
campbell_st
  • campbell_st
it doesn't need L'Hopital ... just factor to find it.... don't over think it
anonymous
  • anonymous
im kinda confuzzled a bit :3
Agent47
  • Agent47
i can do lhopital in my head on this... but i guess it depends
Hero
  • Hero
That's because there's too many people trying to give their version of how to solve it.
anonymous
  • 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
campbell_st
  • campbell_st
looking at the type of questions you need to factorise to find the limit..
Agent47
  • Agent47
abanna, just listen to satellite, he's always right lol
Hero
  • Hero
I object to that statement^
anonymous
  • anonymous
if you get \(\frac{a}{0}\) where \(a\) is some non zero number, then there is no limit
campbell_st
  • campbell_st
@abannavong have a look above at my factorised version.. eliminate the common factor then substitute for your limit.
anonymous
  • anonymous
and if you get \(\frac{0}{0}\) then you know to factor and cancel AND you know how to factor
Hero
  • Hero
I see what you're saying now, @satellite. I usually do that part in my head first, but not on paper.
anonymous
  • anonymous
oh ok
anonymous
  • 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 \(-\frac{1}{2}\)
campbell_st
  • campbell_st
@abannavong what level of maths is this...
anonymous
  • anonymous
and your job is then only to find the other factor, as @campbell_st wrote above
Agent47
  • Agent47
AB calculus, aka cal 1
anonymous
  • anonymous
AP Calc :3
campbell_st
  • campbell_st
ok... so still high school... just I'm in australia...
anonymous
  • anonymous
yeah in grade 12
campbell_st
  • campbell_st
thanks... I think you get 5/8 as the limit... good luck
anonymous
  • anonymous
thank you @campbell_st

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