lim _{x rightarrow 1} (log _3 3x)^{log _x 8} | vedclass

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(A) Let $L = \lim _{x \rightarrow 1} (\log _3 3x)^{\log _x 8}$.This is an indeterminate form of type $1^{\infty}$.We use the formula $\lim _{x \righta

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let $L = \lim _{x \rightarrow 1} (\log _3 3x)^{\log _x 8}$.This is an indeterminate form of type $1^{\infty}$.We use the formula $\lim _{x \rightarrow a} f(x)^{g(x)} = e^{\lim _{x \rightarrow a} g(x)(f(x)-1)}$.Here,$f(x) = \log _3 3x = \log _3 3 + \log _3 x = 1 + \log _3 x$ and $g(x) = \log _x 8$.So,$L = \exp(\lim _{x \rightarrow 1} \log _x 8 (1 + \log _3 x - 1)) = \exp(\lim _{x \rightarrow 1} \log _x 8 \cdot \log _3 x)$.Using the change of base formula,$\log _x 8 = \frac{\ln 8}{\ln x}$ and $\log _3 x = \frac{\ln x}{\ln 3}$.Thus,$L = \exp(\lim _{x \rightarrow 1} \frac{\ln 8}{\ln x} \cdot \frac{\ln x}{\ln 3}) = \exp(\frac{\ln 8}{\ln 3}) = \exp(\log _3 8) = 8$.

$\lim _{x \rightarrow 1} (\log _3 3x)^{\log _x 8} = \ldots$

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