If lim x - 01x8-1-cosx22-cosx24-cosx22cosx24-2-k- then the value of k is-

Lim _x → 01x^8[(1 cosx^22) cosx^24(1 cosx^22)] =lim _x → 01x^8[(1 cosx^22)(1 cosx^24)] =lim _x → 01x^8[2sin ^2x^24 2 sin ^2x^28] =lim_x → 01256(sinx^24(x^24)) ...

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Standard XII

Mathematics

L'Hospital Rule to Remove Indeterminate Form

If lim x → 01...

Question

If $lim x \to 0 \frac{1}{x^{8}} [1 - cos \frac{x^{2}}{2} - cos \frac{x^{2}}{4} + cos \frac{x^{2}}{2} cos \frac{x^{2}}{4}] = 2^{- k},$ then the value of $k$ is

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Solution

$lim x \to 0 \frac{1}{x^{8}} [(1 - cos \frac{x^{2}}{2}) - cos \frac{x^{2}}{4} (1 - cos \frac{x^{2}}{2})]$
$= lim x \to 0 \frac{1}{x^{8}} [(1 - cos \frac{x^{2}}{2}) (1 - cos \frac{x^{2}}{4})]$

$= lim x \to 0 \frac{1}{x^{8}} [2 {sin}^{2} \frac{x^{2}}{4} 2 {sin}^{2} \frac{x^{2}}{8}]$

$= lim x \to 0 \frac{1}{256} {⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{sin \frac{x^{2}}{4}}{(\frac{x^{2}}{4})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠}^{2} {⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{sin \frac{x^{2}}{8}}{(\frac{x^{2}}{8})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠}^{2} = \frac{1}{256} = 2^{- 8}$

$∴ k = 8$

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L'Hospital Rule to Remove Indeterminate Form

Standard XII Mathematics

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If limx→01x8[1−cosx22−cosx24+cosx22cosx24]=2−k, then the value of k is​​​​​​​

limx→01x8[(1−cosx22)−cosx24(1−cosx22)] =limx→01x8[(1−cosx22)(1−cosx24)] =limx→01x8[2sin2x242sin2x28] =limx→01256⎛⎜ ⎜ ⎜ ⎜⎝sinx24(x24)⎞⎟ ⎟ ⎟ ⎟⎠2⎛⎜ ⎜ ⎜ ⎜⎝sinx28(x28)⎞⎟ ⎟ ⎟ ⎟⎠2=1256=2−8 ∴k=8

If $lim x \to 0 \frac{1}{x^{8}} [1 - cos \frac{x^{2}}{2} - cos \frac{x^{2}}{4} + cos \frac{x^{2}}{2} cos \frac{x^{2}}{4}] = 2^{- k},$ then the value of $k$ is