Question

help with AP calc hw plz!

Answer 1

Simply posting your actual question would be great

Answer 2

im going to take a pic :3

Answer 3

the limit is x->-1/2+

Answer 4

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

Answer 5

#38

Answer 6

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

Answer 7

kk

Answer 8

limit as x goes to ???

Answer 9

-1/2

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

@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

Answer 19

so plugin -1/2 im kinda confused

Answer 20

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

Answer 21

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

Answer 22

what i am trying to say is that if i replace \(x\) by \(-\frac{1}{2}\) i know it must factor as
\((2x+1)\times (\text{whatever})\)

Answer 23

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

Answer 24

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

Answer 25

here is the factorised version I hope...

\[\lim_{x \rightarrow -\frac{1}{2}} \frac{(2x + 1)(3x -1)}{(2x + 1)(2x - 3)}\]

Answer 26

kk

Answer 27

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

Answer 28

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

Answer 29

yes, by the "factor theorem"
if \(r\) is a root of a polynomial \(p(x)\) then we know \(p(x)=(x-r)q(x)\)

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

im kinda confuzzled a bit :3

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

I object to that statement^

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

oh ok

Answer 46

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 \(-\frac{1}{2}\)

Answer 47

@abannavong what level of maths is this...

Answer 48

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

Answer 49

AB calculus, aka cal 1

Answer 50

AP Calc :3

Answer 51

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

Answer 52

yeah in grade 12

Answer 53

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

Answer 54

thank you @campbell_st