The value of limx→0(1−12x)(1tan⁡x+4−2) is

=limx→0 2x−12x1tan⁡x+4−2×tan⁡x+4+2tan⁡x+4+2 =limx→0 2x−12xtan⁡x+4+2tan⁡x =limx→0 2x−12xxtan⁡x+4+2tan⁡xx =4log⁡2 =log⁡16

The value of limx→0(1−12x)(1tan⁡x+4−2) is

=limx→0 2x−12x1tan⁡x+4−2×tan⁡x+4+2tan⁡x+4+2 =limx→0 2x−12xtan⁡x+4+2tan⁡x =limx→0 2x−12xxtan⁡x+4+2tan⁡xx =4log⁡2 =log⁡16

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The value of $lim_{x \to 0} (1 - \frac{1}{2^{x}}) (\frac{1}{\sqrt{\tan x + 4} - 2}) is$

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log16

does not exist

3 log2

6 log2

answer is A.

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Detailed Solution

$= lim_{x \to 0} (\frac{2^{x} - 1}{2^{x}}) (\frac{1}{\sqrt{\tan x + 4} - 2} \times \frac{\sqrt{\tan x + 4} + 2}{\sqrt{\tan x + 4} + 2}) = lim_{x \to 0} (\frac{2^{x} - 1}{2^{x}}) (\frac{\sqrt{\tan x + 4} + 2}{\tan x}) = lim_{x \to 0} (\frac{2^{x} - 1}{2^{x} x}) (\frac{\sqrt{\tan x + 4} + 2}{(\frac{\tan x}{x})}) = 4 \log 2 = \log 16$