## baldymcgee6 Group Title Find the value of this limit: one year ago one year ago

1. baldymcgee6 Group Title
2. hartnn Group Title

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

3. baldymcgee6 Group Title

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

4. kulprit Group Title

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

5. baldymcgee6 Group Title

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

6. Jemurray3 Group Title

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

7. kulprit Group Title

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

8. kulprit Group Title

yeah sorry @Jemurray3 limit is -0.5 disregard that

9. baldymcgee6 Group Title

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 :)

10. baldymcgee6 Group Title

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

11. Jemurray3 Group Title

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

12. Jemurray3 Group Title

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

13. Jemurray3 Group Title

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

14. baldymcgee6 Group Title

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

15. baldymcgee6 Group Title

how can we say thay u ->0?

16. Jemurray3 Group Title

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

17. baldymcgee6 Group Title

ooohhh gotcha

18. baldymcgee6 Group Title

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

19. Jemurray3 Group Title

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

20. baldymcgee6 Group Title

is that where we get the -0.5?

21. baldymcgee6 Group Title

oh.. k good. thanks so much

22. Jemurray3 Group Title

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

23. baldymcgee6 Group Title

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

24. Jemurray3 Group Title

right.

25. baldymcgee6 Group Title

thanks, you rock.