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Finding Limits of Trig Functions

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data
    The first problem I'm having difficulty with is

    [itex]\stackrel{lim}{x\rightarrow0}[/itex] [itex]\frac{sin x}{5x}[/itex]

    And the second is:
    [itex]\stackrel{lim}{x\rightarrow0}[/itex] [itex]\frac{sin x(1-cos x)}{2x^{2}}[/itex]

    2. Relevant equations
    I assume that for the first problem I need to simplify it to the rule where [itex]\stackrel{lim}{x\rightarrow0}[/itex] [itex]\frac{sin x}{x}[/itex]=1
    and the second would probably need to simplify to follow the rule [itex]\stackrel{lim}{x\rightarrow0}[/itex] [itex]\frac{1-cos x}{x}[/itex]=0

    3. The attempt at a solution
    What I mainly need help with is how to get started. For the first problem, how do I get rid of the 5x at the bottom?
    For the second problem should I square the entire thing and end up with [itex]\stackrel{lim}{x\rightarrow0}[/itex] [itex]\frac{2sin x}{4x^{4}}[/itex] then go from there or is that even correct?
     
  2. jcsd
  3. Aug 24, 2011 #2

    rock.freak667

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    Homework Helper

    Rewrite the first one as (1/5)(sinx/x) then take the limit.

    use the fact that x2=x*x and then try to get sinx/x and (1-cosx)/x then use the fact that lim(x→a) f*g = lim(x→a) f * lim(x→a) g
     
  4. Aug 25, 2011 #3

    rude man

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    Gold Member

    Use the power series for sin(x) and cos(x).

    You already know the first answer is 1/5.

    Second problem:
    cos x = 1 - ax^2 + bx^4 + ... etc. since it's an even function.
    sin x = cx + dx^3 + .... etc. since it's an odd function.
    Then take the first two terms of cos x and the first term of sin x in your expression, realize that the lowest power of x in the numerator is higher than the highest power of x in the denominator, and bingo!
     
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