Question

`lim_("x" -> 0) ("x cos x" - "log" (1 + "x"))/"x"^2` is equal to ____________.

Options

• 1

• 0

• `1/2`

• None of these

Answer 1

`lim_("x" -> 0) ("x cos x" - "log" (1 + "x"))/"x"^2` is equal to `underline(1/2)`.

Explanation:

`=  lim_("x" -> 0) ("x cos x" - "log" (1 + "x"))/"x"^2`

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

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

`= 1/2` (using L’Hospital Rule)