How can I prove that lim_(x => 0) (x-sin x)/(x^3)=1/6

blackdivcp

blackdivcp

Answered question

2022-10-15

Determine lim x 0 x sin x x 3 = 1 6 , without L'Hospital or Taylor

Answer & Explanation

Davin Meyer

Davin Meyer

Beginner2022-10-16Added 13 answers

Let L = lim x 0 x sin ( x ) x 3 . We then have
L = lim y 0 3 y sin ( 3 y ) 27 y 3 = lim y 0 3 y 3 sin ( y ) + 4 sin 3 ( y ) 27 y 3 sin ( 3 y ) = 3 sin ( y ) 4 sin 3 ( y ) = lim y 0 3 y 3 sin ( y ) 27 y 3 + 4 27 lim y 0 sin 3 ( y ) y 3 = 3 27 L + 4 27
This gives us 24 L = 4 L = 1 6

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