baldymcgee6
Find the value of this limit:
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hartnn
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write x^2-1 as (x+1)(x-1)
baldymcgee6
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Okay, did that, I suspecting something will cancel somewhere?
kulprit
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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 (x-1)*(x+1)
limit will be (1/2)*1 which is 0.5
baldymcgee6
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@kulprit the limit is going to -1 not 0, doesn't that matter?
Jemurray3
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Should be -1/2, I think...
kulprit
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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
kulprit
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yeah sorry @Jemurray3 limit is -0.5
disregard that
baldymcgee6
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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 :)
baldymcgee6
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@Jemurray3, can you help me understand how we can use this trig limit even when x approaches -1 and not 0?
Jemurray3
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you have lim (x->-1) sin(x+1)/(x+1)
Jemurray3
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let u = x+1. As x -> -1, u -> 0, so the above is the same as
lim (u -> 0 ) sin(u)/u
Jemurray3
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should be 1/(x-1) I believe but yes, that's the idea.
baldymcgee6
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\[\frac{ 1 }{ x-1 } \lim_{x \rightarrow -1} \frac{ \sin(x+1) }{ x+1 }\]
baldymcgee6
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how can we say thay u ->0?
Jemurray3
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because (x+1) goes to zero as x goes to -1.
baldymcgee6
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ooohhh gotcha
baldymcgee6
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what do we do with the 1/(x-1)
Jemurray3
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Nothing, that's perfectly continuous as x approaches -1 so that just becomes -1/2.
baldymcgee6
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is that where we get the -0.5?
baldymcgee6
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oh.. k good. thanks so much
Jemurray3
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Oh, I wasn't paying attention... that should be on the right side of the limit. so it should be
\[ \lim_{x \rightarrow -1} \frac{\sin(x+1)}{x^2-1} = \lim_{x \rightarrow -1} \frac{\sin(x+1)}{x+1}\cdot \frac{1}{x-1} \]
\[ = \frac{-1}{2} \]
baldymcgee6
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oh right, because it's not a constant right?
Jemurray3
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right.
baldymcgee6
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thanks, you rock.