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Homework Help: Limits again

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Have a couple of questions actually.

    Problem #1 [tex]\lim_{x\rightarrow 0}{\left(\frac{4x}{\sin 5x}\right)}[/tex]

    2. Relevant equations

    I was told in class to do [tex]\lim_{x\rightarrow 0}{\left(\frac{4x*5}{5\sin 5x}\right)}[/tex]

    3. The attempt at a solution

    so after that I was told that [tex]{\left(\frac{x}{\sin x}\right)} = 1[/tex]
    but my book says [tex]{\left(\frac{x}{\sin x}\right)} = 0[/tex]

    The final answer we were given was [tex]4/5[/tex]

    Could someone explain this to me? I know how he got 4/5, but which sin formula is correct?


    Problem #2 [tex]\lim_{x\rightarrow 0}{\left(\frac{1-\cos x}{x^2}\right)}[/tex]

    I can get to this step:

    [tex]\lim_{x\rightarrow 0}{\left(\frac{\sin^2 x}{\left(1+\cos x\right)x^2}\right)}[/tex]

    but what's next?

    Thanks guys! I have a test tomorrow, so I'm just trying to get some things straight
  2. jcsd
  3. Feb 6, 2008 #2


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    Homework Helper

    as x-->0 sinx/x=1 so that as x-->0 x/sinx =1

    For the second one put

    [tex]\frac{sin^2x}{(1+cosx)x^2}=(\frac{sinx}{x})^2 \frac{1}{1+cosx}[/tex]
    Last edited: Feb 6, 2008
  4. Feb 6, 2008 #3
    x / sin x can become 1 / (sin x / x) which equals 1/1 = 1.
  5. Feb 6, 2008 #4
    well you almost got it, you did the most part, here

    [tex]\lim_{x\rightarrow 0}{\left(\frac{4x*5}{5\sin 5x}\right)}=\frac{4}{5}\lim_{x\rightarrow 0}{\left(\frac{5x}{\sin 5x}\right)}=\frac{4}{5}\lim_{x\rightarrow 0}{\left(\frac{1}{\frac{\{sin 5x}{\ 5x}\right)}}=[/tex] no remember that

    lim(x-->0)sinB/B=1, where B in our case is 5x
  6. Feb 6, 2008 #5
    so [tex]\lim_{x\rightarrow 0}\frac{\sin x}{x}} = \frac{x}{\sin x}} = 1 [/tex]?
  7. Feb 6, 2008 #6
    well yeah, and dont forget the limit sign, otherwise it is not true,lol.
  8. Feb 6, 2008 #7
    you're fast.
    Great, that is all I needed for that one I suppose. Thanks.

    For the second one, are you just factoring out the ^2 so you can get 1?
  9. Feb 6, 2008 #8


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    Homework Helper

    In a basic sense, yes
  10. Feb 6, 2008 #9
    groovy. This one's solved. thanks a bunch guys :D:D:D
  11. Feb 6, 2008 #10
    Another quick question.

    I have [tex]\lim_{x\rightarrow -1}{\left(\frac{\sqrt {x^2 + 8} + 3}{x+1}\right)}[/tex]

    I know you multiply by [tex]\sqrt {x^2 + 8} - 3[/tex] but what if the +3 wasn't there? Or what if it were on the bottom?
  12. Feb 6, 2008 #11
    If it was on the buttom then that limit would be 1, and if it wasnt there then that limit would not exist. Because the two sided limits would not match.
  13. Feb 6, 2008 #12
    Well just for demonstration purposes? Why do we use - 3 but not - 8 ?

    And if it were on the bottom, would it be the same process (assuming there was a limit)
  14. Feb 6, 2008 #13
    The reason why we multiply the bottom and the top by [tex]\sqrt {x^2 + 8} - 3[/tex]
    is because we want to form a difference of squares in the top, that is somethign like this a^2-b^2=(a-b)(a+b),so we can cancele some terms out and thus end up with something that can cancle out with the bottom, this way we manage to facilitate the process.
    Which, if it were in the bottom, the whole expression [tex]\sqrt {x^2 + 8} + 3[/tex]??

    Usually we rationalise either the bottom or the top when we have some simple intermediate forms like 0/0 or infinty/infinity, then sometimes by just rationalizing the bottom or the top we get rid of these intermediate forms, but not always.

    Can you be more specific, what are u really asking?
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