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baldymcgee6

  • 3 years ago

Find the value of this limit:

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  1. baldymcgee6
    • 3 years ago
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    http://gyazo.com/4ad8e332ba4e68323866548f1b84f5bb

  2. hartnn
    • 3 years ago
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    write x^2-1 as (x+1)(x-1)

  3. baldymcgee6
    • 3 years ago
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    Okay, did that, I suspecting something will cancel somewhere?

  4. kulprit
    • 3 years ago
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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

  5. baldymcgee6
    • 3 years ago
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    @kulprit the limit is going to -1 not 0, doesn't that matter?

  6. Jemurray3
    • 3 years ago
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    Should be -1/2, I think...

  7. kulprit
    • 3 years ago
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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

  8. kulprit
    • 3 years ago
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    yeah sorry @Jemurray3 limit is -0.5 disregard that

  9. baldymcgee6
    • 3 years ago
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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 :)

  10. baldymcgee6
    • 3 years ago
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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?

  11. Jemurray3
    • 3 years ago
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    you have lim (x->-1) sin(x+1)/(x+1)

  12. Jemurray3
    • 3 years ago
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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

  13. Jemurray3
    • 3 years ago
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    should be 1/(x-1) I believe but yes, that's the idea.

  14. baldymcgee6
    • 3 years ago
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    \[\frac{ 1 }{ x-1 } \lim_{x \rightarrow -1} \frac{ \sin(x+1) }{ x+1 }\]

  15. baldymcgee6
    • 3 years ago
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    how can we say thay u ->0?

  16. Jemurray3
    • 3 years ago
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    because (x+1) goes to zero as x goes to -1.

  17. baldymcgee6
    • 3 years ago
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    ooohhh gotcha

  18. baldymcgee6
    • 3 years ago
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    what do we do with the 1/(x-1)

  19. Jemurray3
    • 3 years ago
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    Nothing, that's perfectly continuous as x approaches -1 so that just becomes -1/2.

  20. baldymcgee6
    • 3 years ago
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    is that where we get the -0.5?

  21. baldymcgee6
    • 3 years ago
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    oh.. k good. thanks so much

  22. Jemurray3
    • 3 years ago
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    Oh, I wasn't paying attention... that should be on the right side of the limit. so it should be \[ \lim_{x \rightarrow -1} \frac{\sin(x+1)}{x^2-1} = \lim_{x \rightarrow -1} \frac{\sin(x+1)}{x+1}\cdot \frac{1}{x-1} \] \[ = \frac{-1}{2} \]

  23. baldymcgee6
    • 3 years ago
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    oh right, because it's not a constant right?

  24. Jemurray3
    • 3 years ago
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    right.

  25. baldymcgee6
    • 3 years ago
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    thanks, you rock.

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