Question

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

Answer 1

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

Explanation:

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`