YouEvaluate the following limit: limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction.A. limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction equals 10 over 3B. limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction equals 2 over 3C. limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction equals 8 over 3D. limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction equals 0
Question
YouEvaluate the following limit: limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction.A. limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction equals 10 over 3B. limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction equals 2 over 3C. limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction equals 8 over 3D. limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction equals 0
Solution
The limit of the given function as x approaches 0 can be evaluated using L'Hopital's Rule, which states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.
First, let's find the derivative of the numerator and the denominator.
The derivative of the numerator, 2cos(5x) - 2, is -10sin(5x). The derivative of the denominator, 3x, is 3.
Now, we substitute these into our limit:
lim (x->0) [(-10sin(5x))/3]
As x approaches 0, sin(5x) also approaches 0, so the limit is:
-10*0/3 = 0
So, the correct answer is D. limit as x rightwards arrow 0 of fraction numerator 2 cos left parenthesis 5 x right parenthesis minus 2 over denominator 3 x end fraction equals 0.
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