limx 0 0.05em 2^2x-2^x+1+2- cos ⁡2xx^2=

Question

Accepted Answer

Correct answer: B. (b) We have limx 0 0.05em 2^2x-2^x+1+2- cos ⁡2xx^2 at x 0 we get ( 00 ) form using L. Hospital Rule two times, we get 2+ ( log ⁡2 )^2

$lim x \to 0 \frac{2 ^{2 x} - 2 ^{x + 1} + 2 - cos 2 x}{x ^{2}} =$

Solution

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