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sparky34
sparky34 asked in Science & MathematicsMathematics · 8 years ago

lim x->0 (sin(5tan(4x)))/sin(3x)?

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  • Brenda
    Lv 7
    8 years ago
    Favorite Answer

    lim sin[5tan(4x)]/sin(3x)

    x->0

    lim [sin[5tan(4x)]*5tan(4x)/5tan(4x)]/[sin(3x)*3x/3x]

    x->0

    what you can't see after the ... is [sin(3x)*3x/3x]

    lim sin(3x)/3x = 1

    x->0

    lim sin(5tan(4x))/5tan(4x) = 1

    x->0

    so

    lim 5tan(4x)/3x

    x->0

    lim [5sin(4x)/cos(4x)]/3x

    x->0

    lim 5sin(4x)/[3x*cos(4x)]

    x->0

    lim [5sin(4x)*4x/4x]/[3x*cos(4x)]

    x->0

    lim sin(4x)/4x = 1

    x->0

    lim 20x/[3x*cos(4x)]

    x->0

    lim 20/3cos(4x)

    x->0

    as x->0 then 20/3cos(4*0) => 20/3

    The limit is 20/3.

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