Question

Find the value of this limit:

Answer 1

http://gyazo.com/4ad8e332ba4e68323866548f1b84f5bb

Answer 2

write x^2-1 as (x+1)(x-1)

Answer 3

Okay, did that, I suspecting something will cancel somewhere?

Answer 4

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

Answer 5

@kulprit the limit is going to -1 not 0, doesn't that matter?

Answer 6

Should be -1/2, I think...

Answer 7

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

Answer 8

yeah sorry @Jemurray3 limit is -0.5
disregard that

Answer 9

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 :)

Answer 10

@Jemurray3, can you help me understand how we can use this trig limit even when x approaches -1 and not 0?

Answer 11

you have lim (x->-1) sin(x+1)/(x+1)

Answer 12

let u = x+1. As x -> -1, u -> 0, so the above is the same as
lim (u -> 0 ) sin(u)/u

Answer 13

should be 1/(x-1) I believe but yes, that's the idea.

Answer 14

\[\frac{ 1 }{ x-1 } \lim_{x \rightarrow -1} \frac{ \sin(x+1) }{ x+1 }\]

Answer 15

how can we say thay u ->0?

Answer 16

because (x+1) goes to zero as x goes to -1.

Answer 17

ooohhh gotcha

Answer 18

what do we do with the 1/(x-1)

Answer 19

Nothing, that's perfectly continuous as x approaches -1 so that just becomes -1/2.

Answer 20

is that where we get the -0.5?

Answer 21

oh.. k good. thanks so much

Answer 22

Oh, I wasn't paying attention... that should be on the right side of the limit. so it should be
\[ \lim_{x \rightarrow -1} \frac{\sin(x+1)}{x^2-1} = \lim_{x \rightarrow -1} \frac{\sin(x+1)}{x+1}\cdot \frac{1}{x-1} \]
\[ = \frac{-1}{2} \]

Answer 23

oh right, because it's not a constant right?

Answer 24

right.

Answer 25

thanks, you rock.