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baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
http://gyazo.com/4ad8e332ba4e68323866548f1b84f5bb
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
write x^21 as (x+1)(x1)
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
Okay, did that, I suspecting something will cancel somewhere?
 one year ago

kulprit Group TitleBest ResponseYou've already chosen the best response.1
there is an established proof that limit z>0 (sin(z))/z is 1 put z = x+1 z>0 => x>1 write denominator as (x1)*(x+1) limit will be (1/2)*1 which is 0.5
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
@kulprit the limit is going to 1 not 0, doesn't that matter?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Should be 1/2, I think...
 one year ago

kulprit Group TitleBest ResponseYou've already chosen the best response.1
limit z>0 (sin(z))/z is 1 and so limit x>1 (sin(x+1))/(x+1) is 1 just substitute k = x+1 x>1 ,so x+1>0 => k>0
 one year ago

kulprit Group TitleBest ResponseYou've already chosen the best response.1
yeah sorry @Jemurray3 limit is 0.5 disregard that
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
I understand how the lim x>0 sin(x)/x = 1 but I don't understna dhow we can use this when this limit is going to 1, sorry for being difficult :)
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
@Jemurray3, can you help me understand how we can use this trig limit even when x approaches 1 and not 0?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
you have lim (x>1) sin(x+1)/(x+1)
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
let u = x+1. As x > 1, u > 0, so the above is the same as lim (u > 0 ) sin(u)/u
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
should be 1/(x1) I believe but yes, that's the idea.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ 1 }{ x1 } \lim_{x \rightarrow 1} \frac{ \sin(x+1) }{ x+1 }\]
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
how can we say thay u >0?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
because (x+1) goes to zero as x goes to 1.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
ooohhh gotcha
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
what do we do with the 1/(x1)
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Nothing, that's perfectly continuous as x approaches 1 so that just becomes 1/2.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
is that where we get the 0.5?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
oh.. k good. thanks so much
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Oh, I wasn't paying attention... that should be on the right side of the limit. so it should be \[ \lim_{x \rightarrow 1} \frac{\sin(x+1)}{x^21} = \lim_{x \rightarrow 1} \frac{\sin(x+1)}{x+1}\cdot \frac{1}{x1} \] \[ = \frac{1}{2} \]
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
oh right, because it's not a constant right?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
thanks, you rock.
 one year ago
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