lim _{x rightarrow 0} frac{6^x-3^x-2^x+1}{x^2 | vedclass

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(A) Let $l = \lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}$.We can factor the numerator as follows:$6^x - 3^x - 2^x + 1 = 3^x(2^x - 1) - 1(2^x - 1

Vedclass · Accepted Answer

$(\log _e 2)(\log _e 3)$

Vedclass · Answer

(A) Let $l = \lim _{x ightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}$.We can factor the numerator as follows:$6^x - 3^x - 2^x + 1 = 3^x(2^x - 1) - 1(2^x - 1) = (2^x - 1)(3^x - 1)$.Substituting this back into the limit:$l = \lim _{x ightarrow 0} \frac{(2^x - 1)(3^x - 1)}{x^2} = \lim _{x ightarrow 0} \left( \frac{2^x - 1}{x} ight) 	imes \left( \frac{3^x - 1}{x} ight)$.Using the standard limit formula $\lim _{x ightarrow 0} \frac{a^x - 1}{x} = \log _e a$,we get:$l = (\log _e 2) 	imes (\log _e 3)$.

$\lim _{x \rightarrow 0} \frac{6^x-3^x-2^x+1}{x^2}$ is equal to

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