How do you find the limit of $(x-sinx)/ (x^3)$ as x approaches 0?

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Answer 1 · 2017-02-24T09:04:39

$Lt_(x->0)(x-sinx)/x^3=1/6$

Maclaurin series expansion for $sinx$ is given by

$x-x^3/(3!)+x^5/(5!)-x^7/(7!)+x^9/(9!)-..............$

Hence, $(x-sinx)/x^3=1/x^3(x^3/(3!)-x^5/(5!)+x^7/(7!)-x^9/(9!)+........)$

= $1/6-x^2/120+x^4/5040-x^6/362880+........$

and $Lt_(x->0)(x-sinx)/x^3=1/6$

How do you find the limit of (x-sinx)/ (x^3)(x-sinx)/ (x^3) as x approaches 0?