help with AP calc hw plz!

- anonymous

help with AP calc hw plz!

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- Hero

Simply posting your actual question would be great

- anonymous

im going to take a pic :3

- anonymous

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## More answers

- anonymous

the limit is x->-1/2+

- Hero

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

- anonymous

#38

- campbell_st

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

- anonymous

kk

- anonymous

limit as x goes to ???

- Hero

-1/2

- campbell_st

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

- 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

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

- anonymous

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

- anonymous

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

- campbell_st

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

- 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

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

- 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

so plugin -1/2 im kinda confused

- Hero

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

- campbell_st

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

- 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

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

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

- 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

kk

- 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

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

- 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

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

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

- 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

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

- anonymous

im kinda confuzzled a bit :3

- Agent47

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

- Hero

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

- 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

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

- Agent47

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

- Hero

I object to that statement^

- anonymous

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

- campbell_st

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

- anonymous

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

- Hero

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

- anonymous

oh ok

- 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

@abannavong what level of maths is this...

- anonymous

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

- Agent47

AB calculus, aka cal 1

- anonymous

AP Calc :3

- campbell_st

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

- anonymous

yeah in grade 12

- campbell_st

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

- anonymous

thank you @campbell_st

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