sin 3x/7x

i keep getting 0 even using the quotient rule but htis is not the answer

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asked by rudy
  1. the limit as x approaches 0 of (sinx/x) is 1. Hence, you need to get the given expression, sin3x/7x in the form sinu/u. Observe that sin3x/7x = (sin3x/3x)*(3x/7x)... multiply the numerator and denominator by 3x so we have the expression sinu/u. Thus, the limit as x approaches 0 of (sin3x/3x)*(3x/7x)is simply (1)*(3/7).

    You CANNOT use QUOTIENT rule since the question is asking for the VALUE of a particular LIMIT of a function, NOT the slope of the tangent at a particular point. You CAN use L'Hopital's Rule if you know what it is (I suggest you stick with the basics).

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    posted by JP

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