The value of limx - 0 1-8x-9x-72x9x-8xtan 9x - sin 8x is

Lim_x → 0 1 8^x 9^x+(72)^x(9^x 8^x)(tan 9x sin 8x) = lim_x → 0 (9^x 1)(8^x 1)((9^x 1) (8^x 1))(tan 9x sin 8x) =lim_x → 0 9^x 1x8^x 1x(9^x 1x 8^x 1x)(tan 9 x ...

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Byju's Answer

Standard XII

Mathematics

Expansions to Remove Indeterminate Form

The value of ...

Question

The value of $lim x \to 0 \frac{1 - 8^{x} - 9^{x} + (72)^{x}}{(9^{x} - 8^{x}) (tan 9 x - sin 8 x)}$ is

log 8 \cdot log 9 \cdot log (\frac{9}{8})

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log 8 \cdot log 9 \cdot log (\frac{8}{9})

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6 log 8 \cdot log 9 \cdot log (\frac{8}{9})

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\frac{log 8 \cdot log 9}{log (\frac{9}{8})}

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Solution

The correct option is D $\frac{log 8 \cdot log 9}{log (\frac{9}{8})}$
$lim x \to 0 \frac{1 - 8^{x} - 9^{x} + (72)^{x}}{(9^{x} - 8^{x}) (tan 9 x - sin 8 x)}$
$= lim x \to 0 \frac{(9^{x} - 1) (8^{x} - 1)}{((9^{x} - 1) - (8^{x} - 1)) (tan 9 x - sin 8 x)}$
$= lim x \to 0 \frac{\frac{9^{x} - 1}{x} \frac{8^{x} - 1}{x}}{(\frac{9^{x} - 1}{x} - \frac{8^{x} - 1}{x}) (\frac{tan 9 x}{x} - \frac{sin 8 x}{x})}$
$= \frac{ln 9 \cdot ln 8}{(ln 9 - ln 8) (9 - 8)}$
$= \frac{ln 8 \cdot ln 9}{ln (\frac{9}{8})}$

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Similar questions

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f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{(72)^{x} - 9^{x} - 8^{x} + 1}{\sqrt{2} - \sqrt{1 + cos x}} : & x \neq 0 k log 2 log 3; & x = 0 \end{matrix}

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Expansions to Remove Indeterminate Form

Standard XII Mathematics

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The value of limx→01−8x−9x+(72)x(9x−8x)(tan9x−sin8x) is

The correct option is D log8⋅log9log(98)limx→01−8x−9x+(72)x(9x−8x)(tan9x−sin8x) =limx→0(9x−1)(8x−1)((9x−1)−(8x−1))(tan9x−sin8x) =limx→09x−1x8x−1x(9x−1x−8x−1x)(tan9xx−sin8xx) =ln9⋅ln8(ln9−ln8)(9−8) =ln8⋅ln9ln(98)

The value of $lim x \to 0 \frac{1 - 8^{x} - 9^{x} + (72)^{x}}{(9^{x} - 8^{x}) (tan 9 x - sin 8 x)}$ is