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I need AP Calculus help!!

Mathematics
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i will post the assignment soon!
Help with all plz!
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Other answers:

hi
the circled questions btw
for the first one, its just 8 thats posted?
i mean circled
its numbers 10, 14, 16, 18, 22
oh i was looking at the first one
kk
oh man, that second one's really hard to read lol so lets do the first one
kk
k so for 8 what is x approaching? i cant see
we dont have to do number 8 lol
:\ it looked circled to me lol
its only numbers 10, 14, 16, 18, 22
do you not have to do the problem in your first attachment?
oh yeah duh i forgot i do lolz
haha ok cool well i basically cant read any of them cuz they're blurry so can you just type out one that you need help with?
kk
hang on
number 8 its as x goes to infinity
so it's as x-> infinity of (2x^2)/(x^2 +3) ? is that right?
2x^5/ x^5+3
oh ok.
have y'all talked about L'Hospital's Rule yet?
i dont think so
k.
what is the L'hospitals rule? can u plz explain
i don't think you're supposed to use it if you haven't covered it yet. you don't need it anyway.
so you start by considering the terms with the highest exponents
if I remember correctly, since x is approaching infinity and it's a rational function, you can rewrite it only including the x's with the greatest exponents
so what would you have if you did that?
hang on im kinda confused
that's ok. so what i'm saying is your next step is to rewrite the limit excluding the +3 in the denominator
to exclude the +3 in the denominator how do u do tht?
you can do this because when x goes to infinity, 2x^5 and x^5 approach infinity and the 3 basically doesn't change the fraction anymore
because if you imagine really high numbers, say x=100, x^5 would be 10000000000 and 2x^5 would be 20000000000 and the +3 in the denominator makes less and less of a difference in the fraction. (10000000000/20000000003 is really close to 10000000000/20000000000)
oh ok
in other words, the fraction approaches 1/2
im sorry, it approaches 2
because 2x^5 is in the numerator
oh ok so ur limit is basically 2 then right?
yup
oh ok now i get number 8
so in questions like that, where x is approaching infinity, you can disregard any terms that arent the variables with the highest exponent. like if you had an x^4 being added/subtracted somewhere in the fraction, you could disregard that, too.
why can u?
for the same reason, because as x approaches infinity, they change the fraction less and less, even though it has x in it....those terms approach infinity slower than terms of higher powers.
im kinda confused on tht part
it's kind of counter-intuitive at first
oh ok
as x gets greater, the difference between x^5 and x^4 also gets greater. if x is 10, then x^5 is 100000 and x^4 is 10000, x^5 is 10 times greater than x^4. but when x= 100, x^5 becomes 100 times bigger than x^4, and so on, until x^5 is infinitely bigger than x^4 and you can disregard x^4.
oh ok now i get it
:) cool
can u help me with the continuity limits and one-sided ones those ones are the ones that r confusing
i can help w one more then i need to get my own work done haha
kk
so i guess pick one for us to do
kk
i think i get the rest on my THQ but i need help with the continuity ones
k
let me post the problem
kk
\[\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 } \]
so first notice that when x = 4, f(x) = 0/0, which is undefined.
yeah
and since x is approaching 4 from the left (less than 4, getting greater) both the numerator and denominator are going to be negative...do u get that?
can u explain it a bit more im still a bit confused
sure. so the way i think of it, is to basically imagine x as some number a little less than 4.. 3.something. when you plug that number into the denominator, 3.something - 4 is always going to be negative. (anything less than 4) - 4 will be negative, right? (3-4=-1, 3.5-4=-0.5, 3.99-4 = -0.01)
oh yeah so its like it keeps on approaching 4 right?
and for the numerator, you can imagine that since the square root of 4 is 2, the square root of anything less than 4 will be less than 2... when you subtract 2 from that number, the result is negative
ya it keeps approaching 4
but from the less-than-4 side because it has the minus sign (x->4-)
basically the limit for it will negative right?
well the numerator and denominator are both negative, right?
so the fraction comes out positive
oh ok
why r limits soo confusing :3
lol i know. wanna see my limit problem that no one will answer?
Limit as x->(pi/2) of [tanx+1/(x-pi/2)]?
thats my question that no one will answer :(
i had it on a quiz today
can u do the substitution method for tht one
i think
idk
no because tan of pi/2 is undefined, it goes to neg infinity and infinity from the right and left....
oh yeah ur right oops lol
ya...it's brutal.
anyway
but im still kinda confused on the continuity limits and one-sided ones
oh boy satellite's here.... it's about to get real xD
i just realized that your problem is tricky. you need to factor your x-4 into (sqrt(x)-2)(sqrt(x)+2) then cancel out the numerator
\[\lim_{ x\to\frac{\pi}{2}}\frac{\tan(x)+1}{x-\frac{\pi}{2}}\]?
oooh well that's my problem but not hers
mines posted
oh ok
somewhere down the list!
@jldstuff393 how do u do my problem kinda confuzzled
i know i know. u remember difference of squares? like (x^2 - 4) = (x-2)(x+2)?
is this the problem \[\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 }\]?
oh yeah i remember different of squares
yeah @satellite73 thts my problem
so unless satellite has a less tricky way to solve it, you can factor (x-4) as a difference of squares... (sqrt(x)-2)(sqrt(x)+2)
does that make sense?
the gimmick here is to "rationalize the numerator" by multiplying by the conjugate of \(\sqrt{x}-2\) which is \(\sqrt{x}+2\)
cuz when u have that, you can cancel out the (sqrt(x)-2) in the numerator and denominator
so u multpily the top and bottom by \[\sqrt{x} +2 \]
this works because \((a+b)(a-b)=a^2-b^2\) and so \((\sqrt{x}-2)(\sqrt{x}+2)=x-4\)
that cancels with the \(x-4\) in the denominator, which is what you want, you want to cancel the zero
and x-4 can cancel in denominator
oh ok!
so ur limit is one right?
leaves you with \[\lim_{x \rightarrow 4} \frac{ \sqrt{x}-2 }{ x-4 }=\lim_{x \rightarrow 4} \frac{1 }{ \sqrt{x}+2}\]\]
oh ok i get
no limit is not one, now you have to replace \(x\) by 4
once u replace x with 4 u get 1/ 4 right?
yup
whenever you get \(\frac{0}{0}\) you need some gimmick to factor and cancel to get rid of the zeros
satellite would you mind helping me after you're done here? O.o good luck guys!
oh ok
sometime sit is easy sometimes it is a pain. but you get used to it
@satellite73 i have some more questions!
@jldstuff393 sure if i can what is the question
@abannavong go ahead and post them, i will look
i'll repost it, thanks!
kk! thanks @satellite73
14, 16, and 18
i can't really read it, but for 16 replace \(x\) by 2 in both formulas if you get the same number, that is the limit if you get different numbers, there is no limit
number 14 its as deltax goes to 0 with the little plus sign
i can't read 18 either is it \(\lim_{ x\to 1^+}\) or \(\lim_{x\to 1^-}\)?
x-> 1+
then replace \(x\) by 1 in the second formula you get \(1-1=0\)
oh ok
can u help with 14 also
its confusing
sure, but i can't really read it can you post in a new question? you do not need to write \(\Delta x\) you can just write \(h\) instead
it is actually very straight forward we will do a little algebra and that is all
kk hang on dinner quickly
\[(x+h)^2+x+h-(x^2+x)=x^2+2xh+h^2+x+h-x^2-x\] \[\cancel{x^2}+2xh+h^2+\cancel{x}+h-\cancel{x^2}-\cancel{x}=2xh+h+h^2\] divide by \(h\) and get \(2x+1+h\) let \(h\to 0\) be replacing \(h\) by 0 and you get \(2x+1\)
oh ok! @satellite73 im back from dinner
that was a quick meal, you should take your time i wrote the solution above, but i used \(h\) instead of \(\Delta x\) it is mostly just algebra, to get rid of the factor of \(\Delta x\) in the denominator, then you replace it by 0
i did eat dinner
im eating and doing work lolz
you are either eating late or are on the left coast
im on the west coast lolz
after you finish eating, read my solution above i hope all the steps are clear they are almost all algebra steps, only at the end do you replace \(h\) by 0
kk
oh ok its jst basic algebra right? so u foil the (x+h)^2 right?
yes exactly it is \((x+h)(x+h)=x^2+2xh+h^2\)
oh ok and then from there u can combine the like terms and get 2xh+h+h^2
yes,
that is all in the numerator you get \[\frac{2xh+h+h^2}{h}\]
and then can u factor out the h on the numerator
yes factor and cancel or divide each term by \(h\) it is the same thing
oh ok and then u get 2x+1
yes
oh ok! i get tht now
whew that was exhausting
\[\lim_{x \rightarrow2 ^{+}}(2x-\left[ \left| x \right| \right]\] help im confused with this one also
is that an absolute value sign?
or is that the floor function?
its like a bracket and then two straight lines inside
do you know what that function is called?
ughhh i dont remember what it was called
it is either called "floor" or "greatest integer"
is it kinda like a piecewise function or something like tht
oh yeah i remember!
so if \(x>2\) but close to 2, then \([x]=2\)
so you have \[\lim_{x\to 2}2x-2=4-2=2\]
to be more precise \[\lim_{x\to 2^+}2x-2=4-2=2\]
oh ok! at first it seemed confusing but oh ok now i get it!
whereas \[\lim_{x\to 2^-}2x-2=4-1=3\] because if \(x<2\) but close to 2, then \([x]=1\)
oh ok! tht seems soo much easier now!
yes if you know what it means it is not hard here is a nice picture of \(y=2x-[x]\) http://www.wolframalpha.com/input/?i=2x-floor%28x%29
oh ok! i get it! i have another question again hang on
so it says find the x-values(if any) at which f is not continuous. which of the discontinuities are removable. f(x)= 1/ x^2+1
\[f(x)=\frac{1}{x^2+1}\] has no discontinuities, since it is a rational function and the denominator is never 0 since \(x^2\geq 0\) for any \(x\) you can see that \(x^2+1\geq 1\) no matter what \(x\) is, so this function has no discontinuities. as the denominator is never zero
oh ok thank you @satellite73
you are welcome enough math, study something else or do something more constructive
pretty much tht was my last problem! time to find a prayer for theology class! thank you @satellite73 and study for english test tomorrow on grapes of wrath
good luck with the joads
lolz thanks i have to find the intercalary chapters to write about :3 i dont even know what those r

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