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HeroBest ResponseYou've already chosen the best response.0
Simply posting your actual question would be great
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
im going to take a pic :3
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
the limit is x>1/2+
 one year ago

HeroBest ResponseYou've already chosen the best response.0
You've got like 5 questions circled. Which one is it?
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
perhaps factorising the numerator and denominator will allow you to cancel a common factor
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
limit as x goes to ???
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
there is a common factor that is evident when you factorise numerator and denominator
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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
 one year ago

HeroBest ResponseYou've already chosen the best response.0
I think @campbell_st is right. You should try factoring the numerator and denominator first before plugging in x = 1/2
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
if you get \(\frac{2}{0}\) then you know you can factor both top and bottom as \((2x+1)(\text{something)}\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
i wouldn't do that, i would plug in first
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
the other thing about eliminating a common factor is that it makes the substitution calculation easier
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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
 one year ago

HeroBest ResponseYou've already chosen the best response.0
If you plug in first, you're definitely going to get a denominator you won't like
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
@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
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
so plugin 1/2 im kinda confused
 one year ago

HeroBest ResponseYou've already chosen the best response.0
Why waste your time trying to evaluate before factoring? Seems silly
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
lol... people get lost with squaring and substituting negative fractions... its really what works best you each of us
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
what i am trying to say is that if i replace \(x\) by \(\frac{1}{2}\) i know it must factor as \((2x+1)\times (\text{whatever})\)
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
because it tells me exactly how to factor otherwise i am doing the dance of what two numbers blah blah blah
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
here is the factorised version I hope... \[\lim_{x \rightarrow \frac{1}{2}} \frac{(2x + 1)(3x 1)}{(2x + 1)(2x  3)}\]
 one year ago

HeroBest ResponseYou've already chosen the best response.0
@satellite lost me. You're telling me that plugging in x = 1/2 tells you which two numbers you use to factor it?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
@hero both @zarkon and i answered the question "why evaluate first" i think
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
yes, by the "factor theorem" if \(r\) is a root of a polynomial \(p(x)\) then we know \(p(x)=(xr)q(x)\)
 one year ago

HeroBest ResponseYou've already chosen the best response.0
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
 one year ago

Agent47Best ResponseYou've already chosen the best response.0
This limit is begging for L'Hopital's Rule.....
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
it doesn't need L'Hopital ... just factor to find it.... don't over think it
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
im kinda confuzzled a bit :3
 one year ago

Agent47Best ResponseYou've already chosen the best response.0
i can do lhopital in my head on this... but i guess it depends
 one year ago

HeroBest ResponseYou've already chosen the best response.0
That's because there's too many people trying to give their version of how to solve it.
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
looking at the type of questions you need to factorise to find the limit..
 one year ago

Agent47Best ResponseYou've already chosen the best response.0
abanna, just listen to satellite, he's always right lol
 one year ago

HeroBest ResponseYou've already chosen the best response.0
I object to that statement^
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
if you get \(\frac{a}{0}\) where \(a\) is some non zero number, then there is no limit
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
@abannavong have a look above at my factorised version.. eliminate the common factor then substitute for your limit.
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
and if you get \(\frac{0}{0}\) then you know to factor and cancel AND you know how to factor
 one year ago

HeroBest ResponseYou've already chosen the best response.0
I see what you're saying now, @satellite. I usually do that part in my head first, but not on paper.
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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 \(\frac{1}{2}\)
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
@abannavong what level of maths is this...
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
and your job is then only to find the other factor, as @campbell_st wrote above
 one year ago

Agent47Best ResponseYou've already chosen the best response.0
AB calculus, aka cal 1
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
ok... so still high school... just I'm in australia...
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
thanks... I think you get 5/8 as the limit... good luck
 one year ago

abannavongBest ResponseYou've already chosen the best response.0
thank you @campbell_st
 one year ago
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