Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

help with AP calc hw plz!

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

Simply posting your actual question would be great
im going to take a pic :3

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question