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Hero
 2 years ago
Best ResponseYou've already chosen the best response.0Simply posting your actual question would be great

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0im going to take a pic :3

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0the limit is x>1/2+

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0You've got like 5 questions circled. Which one is it?

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2limit as x goes to ???

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2then 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
 2 years ago
Best ResponseYou've already chosen the best response.0I think @campbell_st is right. You should try factoring the numerator and denominator first before plugging in x = 1/2

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2if you get \(\frac{2}{0}\) then you know you can factor both top and bottom as \((2x+1)(\text{something)}\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2i wouldn't do that, i would plug in first

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2if 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
 2 years ago
Best ResponseYou've already chosen the best response.0If you plug in first, you're definitely going to get a denominator you won't like

satellite73
 2 years ago
Best 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

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0so plugin 1/2 im kinda confused

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0Why waste your time trying to evaluate before factoring? Seems silly

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2what 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
 2 years ago
Best ResponseYou've already chosen the best response.1rational 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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2because it tells me exactly how to factor otherwise i am doing the dance of what two numbers blah blah blah

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

Hero
 2 years ago
Best 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?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2@hero both @zarkon and i answered the question "why evaluate first" i think

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2yes, by the "factor theorem" if \(r\) is a root of a polynomial \(p(x)\) then we know \(p(x)=(xr)q(x)\)

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0No, @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
 2 years ago
Best ResponseYou've already chosen the best response.0This limit is begging for L'Hopital's Rule.....

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2it 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
 2 years ago
Best ResponseYou've already chosen the best response.2it doesn't need L'Hopital ... just factor to find it.... don't over think it

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0im kinda confuzzled a bit :3

Agent47
 2 years ago
Best ResponseYou've already chosen the best response.0i can do lhopital in my head on this... but i guess it depends

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0That's because there's too many people trying to give their version of how to solve it.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2what 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
 2 years ago
Best ResponseYou've already chosen the best response.2looking at the type of questions you need to factorise to find the limit..

Agent47
 2 years ago
Best ResponseYou've already chosen the best response.0abanna, just listen to satellite, he's always right lol

Hero
 2 years ago
Best ResponseYou've already chosen the best response.0I object to that statement^

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2if you get \(\frac{a}{0}\) where \(a\) is some non zero number, then there is no limit

campbell_st
 2 years ago
Best 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.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2and if you get \(\frac{0}{0}\) then you know to factor and cancel AND you know how to factor

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

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2because 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
 2 years ago
Best ResponseYou've already chosen the best response.2@abannavong what level of maths is this...

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.2and your job is then only to find the other factor, as @campbell_st wrote above

campbell_st
 2 years ago
Best ResponseYou've already chosen the best response.2ok... so still high school... just I'm in australia...

campbell_st
 2 years ago
Best ResponseYou've already chosen the best response.2thanks... I think you get 5/8 as the limit... good luck

abannavong
 2 years ago
Best ResponseYou've already chosen the best response.0thank you @campbell_st
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