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The value of limx→0{sinx-x+x36x5} is 1k, then k is ______.

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Question

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

Options

  • 110

  • 120

  • 130

  • 140

MCQ
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Solution

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

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Limits Using L-hospital's Rule
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