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baldymcgee6Best ResponseYou've already chosen the best response.0
http://gyazo.com/4ad8e332ba4e68323866548f1b84f5bb
 one year ago

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

baldymcgee6Best ResponseYou've already chosen the best response.0
Okay, did that, I suspecting something will cancel somewhere?
 one year ago

kulpritBest 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

baldymcgee6Best ResponseYou've already chosen the best response.0
@kulprit the limit is going to 1 not 0, doesn't that matter?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
Should be 1/2, I think...
 one year ago

kulpritBest 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

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

baldymcgee6Best 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

baldymcgee6Best 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

Jemurray3Best ResponseYou've already chosen the best response.1
you have lim (x>1) sin(x+1)/(x+1)
 one year ago

Jemurray3Best 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

Jemurray3Best ResponseYou've already chosen the best response.1
should be 1/(x1) I believe but yes, that's the idea.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
\[\frac{ 1 }{ x1 } \lim_{x \rightarrow 1} \frac{ \sin(x+1) }{ x+1 }\]
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
how can we say thay u >0?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
because (x+1) goes to zero as x goes to 1.
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
what do we do with the 1/(x1)
 one year ago

Jemurray3Best 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

baldymcgee6Best ResponseYou've already chosen the best response.0
is that where we get the 0.5?
 one year ago

baldymcgee6Best ResponseYou've already chosen the best response.0
oh.. k good. thanks so much
 one year ago

Jemurray3Best 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

baldymcgee6Best ResponseYou've already chosen the best response.0
oh right, because it's not a constant right?
 one year ago
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