abannavong
I need AP Calculus help!!
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abannavong
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i will post the assignment soon!
abannavong
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Help with all plz!
abannavong
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jldstuff393
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hi
abannavong
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the circled questions btw
jldstuff393
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for the first one, its just 8 thats posted?
jldstuff393
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i mean circled
abannavong
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its numbers 10, 14, 16, 18, 22
jldstuff393
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oh i was looking at the first one
abannavong
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kk
jldstuff393
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oh man, that second one's really hard to read lol so lets do the first one
abannavong
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kk
jldstuff393
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k so for 8 what is x approaching? i cant see
abannavong
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we dont have to do number 8 lol
jldstuff393
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:\ it looked circled to me lol
abannavong
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its only numbers 10, 14, 16, 18, 22
jldstuff393
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do you not have to do the problem in your first attachment?
abannavong
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oh yeah duh i forgot i do lolz
jldstuff393
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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?
abannavong
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kk
abannavong
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hang on
abannavong
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number 8 its as x goes to infinity
jldstuff393
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so it's as x-> infinity of (2x^2)/(x^2 +3) ? is that right?
abannavong
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2x^5/ x^5+3
jldstuff393
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oh ok.
jldstuff393
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have y'all talked about L'Hospital's Rule yet?
abannavong
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i dont think so
jldstuff393
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k.
abannavong
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what is the L'hospitals rule? can u plz explain
jldstuff393
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i don't think you're supposed to use it if you haven't covered it yet. you don't need it anyway.
jldstuff393
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so you start by considering the terms with the highest exponents
jldstuff393
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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
jldstuff393
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so what would you have if you did that?
abannavong
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hang on im kinda confused
jldstuff393
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that's ok. so what i'm saying is your next step is to rewrite the limit excluding the +3 in the denominator
abannavong
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to exclude the +3 in the denominator how do u do tht?
jldstuff393
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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
jldstuff393
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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)
abannavong
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oh ok
jldstuff393
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in other words, the fraction approaches 1/2
jldstuff393
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im sorry, it approaches 2
jldstuff393
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because 2x^5 is in the numerator
abannavong
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oh ok so ur limit is basically 2 then right?
jldstuff393
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yup
abannavong
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oh ok now i get number 8
jldstuff393
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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.
abannavong
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why can u?
jldstuff393
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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.
abannavong
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im kinda confused on tht part
jldstuff393
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it's kind of counter-intuitive at first
abannavong
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oh ok
jldstuff393
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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.
abannavong
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oh ok now i get it
jldstuff393
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:) cool
abannavong
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can u help me with the continuity limits and one-sided ones those ones are the ones that r confusing
jldstuff393
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i can help w one more then i need to get my own work done haha
abannavong
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kk
jldstuff393
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so i guess pick one for us to do
abannavong
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kk
abannavong
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i think i get the rest on my THQ but i need help with the continuity ones
jldstuff393
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k
abannavong
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let me post the problem
jldstuff393
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kk
abannavong
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\[\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 } \]
jldstuff393
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so first notice that when x = 4, f(x) = 0/0, which is undefined.
abannavong
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yeah
jldstuff393
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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?
abannavong
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can u explain it a bit more im still a bit confused
jldstuff393
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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)
abannavong
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oh yeah so its like it keeps on approaching 4 right?
jldstuff393
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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
jldstuff393
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ya it keeps approaching 4
jldstuff393
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but from the less-than-4 side because it has the minus sign (x->4-)
abannavong
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basically the limit for it will negative right?
jldstuff393
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well the numerator and denominator are both negative, right?
jldstuff393
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so the fraction comes out positive
abannavong
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oh ok
abannavong
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why r limits soo confusing :3
jldstuff393
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lol i know. wanna see my limit problem that no one will answer?
jldstuff393
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Limit as x->(pi/2) of [tanx+1/(x-pi/2)]?
jldstuff393
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thats my question that no one will answer :(
jldstuff393
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i had it on a quiz today
abannavong
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can u do the substitution method for tht one
abannavong
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i think
abannavong
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idk
jldstuff393
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no because tan of pi/2 is undefined, it goes to neg infinity and infinity from the right and left....
abannavong
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oh yeah ur right oops lol
jldstuff393
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ya...it's brutal.
jldstuff393
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anyway
abannavong
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but im still kinda confused on the continuity limits and one-sided ones
nj1202
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oh boy satellite's here.... it's about to get real xD
jldstuff393
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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
anonymous
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\[\lim_{ x\to\frac{\pi}{2}}\frac{\tan(x)+1}{x-\frac{\pi}{2}}\]?
jldstuff393
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oooh well that's my problem but not hers
jldstuff393
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mines posted
anonymous
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oh ok
jldstuff393
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somewhere down the list!
abannavong
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@jldstuff393 how do u do my problem kinda confuzzled
jldstuff393
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i know i know. u remember difference of squares? like (x^2 - 4) = (x-2)(x+2)?
anonymous
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is this the problem
\[\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 }\]?
abannavong
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oh yeah i remember different of squares
abannavong
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yeah @satellite73 thts my problem
jldstuff393
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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)
jldstuff393
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does that make sense?
anonymous
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the gimmick here is to "rationalize the numerator" by multiplying by the conjugate of \(\sqrt{x}-2\) which is \(\sqrt{x}+2\)
jldstuff393
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cuz when u have that, you can cancel out the (sqrt(x)-2) in the numerator and denominator
abannavong
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so u multpily the top and bottom by \[\sqrt{x} +2 \]
anonymous
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this works because \((a+b)(a-b)=a^2-b^2\) and so \((\sqrt{x}-2)(\sqrt{x}+2)=x-4\)
anonymous
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that cancels with the \(x-4\) in the denominator, which is what you want, you want to cancel the zero
abannavong
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and x-4 can cancel in denominator
abannavong
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oh ok!
abannavong
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so ur limit is one right?
anonymous
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leaves you with
\[\lim_{x \rightarrow 4} \frac{ \sqrt{x}-2 }{ x-4 }=\lim_{x \rightarrow 4} \frac{1 }{ \sqrt{x}+2}\]\]
abannavong
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oh ok i get
anonymous
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no limit is not one, now you have to replace \(x\) by 4
abannavong
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once u replace x with 4 u get 1/ 4 right?
anonymous
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yup
anonymous
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whenever you get \(\frac{0}{0}\) you need some gimmick to factor and cancel to get rid of the zeros
jldstuff393
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satellite would you mind helping me after you're done here? O.o good luck guys!
abannavong
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oh ok
anonymous
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sometime sit is easy
sometimes it is a pain.
but you get used to it
abannavong
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@satellite73 i have some more questions!
anonymous
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@jldstuff393 sure if i can
what is the question
anonymous
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@abannavong go ahead and post them, i will look
jldstuff393
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i'll repost it, thanks!
abannavong
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kk! thanks @satellite73
abannavong
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14, 16, and 18
anonymous
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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
abannavong
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number 14 its as deltax goes to 0 with the little plus sign
anonymous
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i can't read 18 either
is it \(\lim_{ x\to 1^+}\) or \(\lim_{x\to 1^-}\)?
abannavong
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x-> 1+
anonymous
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then replace \(x\) by 1 in the second formula
you get \(1-1=0\)
abannavong
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oh ok
abannavong
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can u help with 14 also
abannavong
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its confusing
anonymous
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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
anonymous
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it is actually very straight forward
we will do a little algebra and that is all
abannavong
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kk hang on dinner quickly
anonymous
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\[(x+h)^2+x+h-(x^2+x)=x^2+2xh+h^2+x+h-x^2-x\]
\[\cancel{x^2}+2xh+h^2+\cancel{x}+h-\cancel{x^2}-\cancel{x}=2xh+h+h^2\]
divide by \(h\) and get \(2x+1+h\)
let \(h\to 0\) be replacing \(h\) by 0 and you get \(2x+1\)
abannavong
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oh ok! @satellite73 im back from dinner
anonymous
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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
abannavong
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i did eat dinner
abannavong
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im eating and doing work lolz
anonymous
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you are either eating late or are on the left coast
abannavong
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im on the west coast lolz
anonymous
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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
abannavong
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kk
abannavong
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oh ok its jst basic algebra right? so u foil the (x+h)^2 right?
anonymous
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yes exactly
it is \((x+h)(x+h)=x^2+2xh+h^2\)
abannavong
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oh ok and then from there u can combine the like terms and get 2xh+h+h^2
anonymous
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yes,
anonymous
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that is all in the numerator
you get
\[\frac{2xh+h+h^2}{h}\]
abannavong
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and then can u factor out the h on the numerator
anonymous
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yes factor and cancel or divide each term by \(h\) it is the same thing
abannavong
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oh ok and then u get 2x+1
anonymous
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yes
abannavong
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oh ok! i get tht now
anonymous
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whew that was exhausting
abannavong
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\[\lim_{x \rightarrow2 ^{+}}(2x-\left[ \left| x \right| \right]\] help im confused with this one also
anonymous
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is that an absolute value sign?
anonymous
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or is that the floor function?
abannavong
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its like a bracket and then two straight lines inside
anonymous
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do you know what that function is called?
abannavong
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ughhh i dont remember what it was called
anonymous
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it is either called "floor" or "greatest integer"
abannavong
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is it kinda like a piecewise function or something like tht
abannavong
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oh yeah i remember!
anonymous
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so if \(x>2\) but close to 2, then \([x]=2\)
anonymous
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so you have
\[\lim_{x\to 2}2x-2=4-2=2\]
anonymous
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to be more precise
\[\lim_{x\to 2^+}2x-2=4-2=2\]
abannavong
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oh ok! at first it seemed confusing but oh ok now i get it!
anonymous
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whereas
\[\lim_{x\to 2^-}2x-2=4-1=3\] because if \(x<2\) but close to 2, then \([x]=1\)
abannavong
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oh ok! tht seems soo much easier now!
abannavong
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oh ok! i get it! i have another question again hang on
abannavong
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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
anonymous
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\[f(x)=\frac{1}{x^2+1}\] 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
abannavong
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oh ok thank you @satellite73
anonymous
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you are welcome
enough math, study something else
or do something more constructive
abannavong
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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
anonymous
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good luck with the joads
abannavong
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lolz thanks i have to find the intercalary chapters to write about :3 i dont even know what those r