Question

`lim_("x" -> 0) (2  "sin x - sin"  2 "x")/"x"^3` is equal to ____________.

Options

• None of these

• 1

• 0

• `1/2`

Answer 1

`lim_("x" -> 0) (2  "sin x - sin"  2 "x")/"x"^3` is equal to 1.

Explanation:

`lim_("x" -> 0) (2  "sin x - sin"  2 "x")/"x"^3`

`= lim_("x" -> 0) (2  "sin x" (1 - "cos x"))/"x"^3`

`= 2  lim _("x" -> 0) "sin x"/"x". (1 - "cos x")/"x"^3`

`= 2 xx 1 xx 1/2 = 1`