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Limits next question

  1. Jul 6, 2009 #1
    [tex]\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}[/tex]

    so this is the question.

    I'm here solving this problem you please check where am i wrong or next idea i've to use here.
    =[tex]\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}x\frac{cos{x}+cos{y}}{cos{x}+cos{y}}[/tex]
    =[tex]\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{{x-y}{cos{x}+cos{y}}[/tex]
    =[tex]\lim_{x\rightarrowy}\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x-y}}{{x-y}{cos{x}+cos{y}}[/tex] +\fracsin{x-y}{(x-y)(cos{x}+cos{y})}[/tex]

    after this i don't 've idea wht to do.Is there next idea we have to include overhere?
     
  2. jcsd
  3. Jul 6, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The "x" you had in the middle here was just "times" wasn't it? Better not to use such a symbol along with"x" as a variable.

    [tex]=[tex]\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{(x-y)(cos{x}+cos{y})}[/tex]
    =[tex]\lim_{x\rightarrow y}\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x-y}}{(x-y)(cos{x}+cos{y})} +\frac{sin(x-y)}{(x-y)(cos{x}+cos{y})}[/tex]

    after this i don't 've idea wht to do.Is there next idea we have to include overhere?[/QUOTE]
    I think I have corrected your LaTex properly. sin(x- y)= sin(x)cos(y)- cos(x)sin(y) and you appear to be trying to put the numerator in that form. (It would be a good idea to explain things like that when asking about a problem.) You wind up with two parts:
    [tex]\frac{sin x cos y- cos x sin y}{x- y}= \frac{sin(x-y)}{x- y}[/tex]
    which goes to 1 as x goes to y, and
    [tex]\frac{sin x cos x- sin y cos y}{x- y}[/tex]
    which is still a problem.

    It would be far simpler to use L'Hopital's rule. Are you allowed to do that?
     
  4. Jul 7, 2009 #3
    [tex]\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}[/tex]

    so this is the question.

    I'm here solving this problem you please check where am i wrong or next idea i've to use here.
    =[tex]\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}*\frac{cos{x}+cos{y}}{cos{x}+cos{y}}[/tex]
    =[tex]\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{(x-y)(cos{x}+cos{y})}[/tex]
    =[tex]\lim_{x\rightarrowy}\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x-y}}{(x-y)(cos{x}+cos{y})} +\fracsin{x-y}{(x-y)(cos{x}+cos{y})}[/tex]

    after this i don't 've idea wht to do.Is there next idea we have to include over
    here?
     
  5. Jul 7, 2009 #4
    I think I have corrected your LaTex properly. sin(x- y)= sin(x)cos(y)- cos(x)sin(y) and you appear to be trying to put the numerator in that form. (It would be a good idea to explain things like that when asking about a problem.) You wind up with two parts:
    [tex]\frac{sin x cos y- cos x sin y}{x- y}= \frac{sin(x-y)}{x- y}[/tex]
    which goes to 1 as x goes to y, and
    [tex]\frac{sin x cos x- sin y cos y}{x- y}[/tex]
    which is still a problem.

    It would be far simpler to use L'Hopital's rule. Are you allowed to do that?[/QUOTE]

    Yea that's what i meant to express in latex form.Thanks for your correction.

    Isn't it solved by that method?[method which i did]

    Umm..L'Hopital's rule is it the best way to solve the problem?Thanks I don't 've idea bout that but i can get it after reading this rule.If not i'll again ask you.
     
  6. Jul 9, 2009 #5

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Laven! :smile:

    You need to learn your trigonometric identities …

    in this case sinx - siny = 2 sin((x - y)/2) cos((x + y)/2) :wink:
     
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