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March 28, 2017

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find the lim (cos x-1)/ (sin 2x)
x->0

  • maths - ,

    can you use L'Hopital's rule?

    if so
    lim (cos x-1)/ (sin 2x)
    x->0
    = lim (-sinx)/(2cos 2x) as x ---> 0
    = 0/2 = 0

  • maths - ,

    no u cant

  • maths - ,

    some common trig limits you should have come across:
    lim( (sinx)/x ) as x --> 0 = 1
    lim( x/sinx ) as x --> 0 = 1
    lim( (cosx - 1)/x) as x -- 0 = 0

    so lets rewrite the question as

    lim[ ((cos x-1)/x) * x(2sinxcosx) ]
    = lim( (cosx - 1)/x) * lim (x/sinx) * lim (1/2cosx)
    = 0(1)(1/2) = 0

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