Question

The value of `lim_(x→0){(sinx - x + x^3/6)/x^5}` is `1/k`, then k is ______.

पर्याय

• 110

• 120

• 130

• 140

Answer 1

The value of `lim_(x→0){(sinx - x + x^3/6)/x^5}` is `1/k`, then k is 120.

Explanation:

Using L–Hospital's rule,

`lim_(x→0){(sinx - x + x^3/6)/x^5} = lim_(x→∞) (cosx - 1 + (3x^2)/6)/(5x^4)`

= `lim_(x→0)(-sinx + (6x)/6)/(20x^3)`

= `lim_(x→∞)(-cosx + 1)/(60x^2)`

= `lim_(x→0) sinx/(120x)`

= `lim_(x→∞) cosx/120` = `1/120`

 ⇒ k = 120