How can I evaluate $lim_(x->0) (sinx-x)/x^3$ without using L'Hopital's rule?

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Answer 1 · 2017-12-23T17:51:17

$-1/6$ .

Suppose that $"the Reqd. Limit L="lim_(x to 0)(sinx-x)/x^3$ .

Substiture $x=3y," so that, as "x to 0, y to 0$ .

$:. L=lim_(y to 0)(sin3y-3y)/((3y)^3)$ ,

$=lim_(y to 0){(3siny-4sin^3y)-3y}/(27y^3)$ ,

$=lim_(y to 0){(3(siny-y))/(27y^3)-(4sin^3y)/(27y^3)}$ ,

$rArr L=lim_(y to 0)1/9*((siny-y)/y^3)-4/27*(siny/y)^3...(ast)$ .

Note that, here,

$lim_(y to 0)((siny-y)/y^3)=lim_(x to 0)((sinx-x)/x^3)=L$ .

Therefore, $(ast) rArr L=1/9*L-4/27, or, 8/9L=-4/27$ .

Hence, $L=-4/27*9/8=-1/6$ .

Enjoy Maths.!

Answer 2 · 2017-12-24T08:54:40

Kindly refer to the Explanation.

In this Second Method, we use the following series :

$sinx=x-x^3/3!+x^5/5!-...oo$ .

$:. sinx -x=-x^3/3!+x^5/5!-x^7/7!+...oo$ .

$:. (sinx-x)/x^3=-1/6+x^2/5!-x^5/7!+...oo$ .

$rArr lim_(x to 0)(sinx-x)/x^3=-1/6+0-0+...oo$

$i.e., lim_(x to 0)(sinx-x)/x^3=-1/6,$ as in Method 1.

How can I evaluate lim_(x->0) (sinx-x)/x^3lim_(x->0) (sinx-x)/x^3 without using L'Hopital's rule?