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http://gyazo.com/4ad8e332ba4e68323866548f1b84f5bb
write x^2-1 as (x+1)(x-1)
Okay, did that, I suspecting something will cancel somewhere?

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Other answers:

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

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