Question

Lim x->0 (tan x- sin x)/x^3=?

a. 0
b. 1
c. 1/2
d. -1
e. -1/2

Answer 1

`lim_(x->0) (tanx-sinx)/x^3 = 1/2`

Explanation:

Transform the function in this way:

`(tanx-sinx)/x^3 = 1/x^3(sinx/cosx-sinx)`

`(tanx-sinx)/x^3 = 1/x^3((sinx-sinxcosx)/cosx)`

`(tanx-sinx)/x^3 = sinx/x^3(1-cosx)/cosx`

`(tanx-sinx)/x^3 = (sinx/x)((1-cosx)/x^2)( 1/cosx)`

We can use now the well known trigonometric limit:

`lim_(x->0) sinx / x = 1`

and using the trigonometric identity:

`sin^2alpha = (1-cos2alpha)/2`

we have:

`lim_(x->0) (1-cosx)/x^2 = lim_(x->0) (2sin^2(x/2))/x^2 = 1/2 lim_(x->0) (sin(x/2)/(x/2))^2 = 1/2`

While the third function is continuous so:

`lim_(x->0) 1/cosx = 1/1 = 1`

and we can conclude that:

`lim_(x->0) (tanx-sinx)/x^3 = lim_(x->0) (sinx/x)((1-cosx)/x^2)( 1/cosx) = 1 xx 1/2 xx 1 = 1/2`

graph{(tanx-sinx)/x^3 [-1.25, 1.25, -0.025, 1]}