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baldymcgee6
 3 years ago
Find the value of this limit:
baldymcgee6
 3 years ago
Find the value of this limit:

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hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0write x^21 as (x+1)(x1)

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, did that, I suspecting something will cancel somewhere?

kulprit
 3 years ago
Best ResponseYou've already chosen the best response.1there 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

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0@kulprit the limit is going to 1 not 0, doesn't that matter?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Should be 1/2, I think...

kulprit
 3 years ago
Best ResponseYou've already chosen the best response.1limit 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

kulprit
 3 years ago
Best ResponseYou've already chosen the best response.1yeah sorry @Jemurray3 limit is 0.5 disregard that

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0I 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 :)

baldymcgee6
 3 years ago
Best 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?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1you have lim (x>1) sin(x+1)/(x+1)

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

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1should be 1/(x1) I believe but yes, that's the idea.

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{ x1 } \lim_{x \rightarrow 1} \frac{ \sin(x+1) }{ x+1 }\]

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0how can we say thay u >0?

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1because (x+1) goes to zero as x goes to 1.

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0what do we do with the 1/(x1)

Jemurray3
 3 years ago
Best ResponseYou've already chosen the best response.1Nothing, that's perfectly continuous as x approaches 1 so that just becomes 1/2.

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0is that where we get the 0.5?

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0oh.. k good. thanks so much

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

baldymcgee6
 3 years ago
Best ResponseYou've already chosen the best response.0oh right, because it's not a constant right?
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