lim _{x rightarrow 0} frac{2^{2 x}-2^{x+1}+2- | vedclass

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(B) दिया गया है,$\lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2}$ $= \lim _{x \rightarrow 0} \frac{(2^x)^2 - 2 \cdot 2^x + 1 + 1 - \cos

Vedclass · Accepted Answer

$2+(\log _e 2)^2$

Vedclass · Answer

(B) दिया गया है,$\lim _{x \rightarrow 0} \frac{2^{2 x}-2^{x+1}+2-\cos 2 x}{x^2}$ $= \lim _{x \rightarrow 0} \frac{(2^x)^2 - 2 \cdot 2^x + 1 + 1 - \cos 2x}{x^2}$ $= \lim _{x \rightarrow 0} \frac{(2^x - 1)^2 + (1 - \cos 2x)}{x^2}$ $= \lim _{x \rightarrow 0} \left( \frac{2^x - 1}{x} \right)^2 + \lim _{x \rightarrow 0} \frac{1 - \cos 2x}{x^2}$ मानक सीमाओं $\lim _{x \rightarrow 0} \frac{a^x - 1}{x} = \log _e a$ और $\lim _{x \rightarrow 0} \frac{1 - \cos kx}{x^2} = \frac{k^2}{2}$ का उपयोग करने पर: $= (\log _e 2)^2 + \frac{2^2}{2}$ $= (\log _e 2)^2 + 2$

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

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