If lim _{n rightarrow infty} x^n log _e x=0,t | vedclass

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(A) Given,$\lim _{n \rightarrow \infty} x^n \log _e x=0$. We know that as $n \rightarrow \infty$,$x^n \rightarrow 0$ only when $x \in (0, 1)$. If $x >

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Negative

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(A) Given,$\lim _{n \rightarrow \infty} x^n \log _e x=0$. We know that as $n \rightarrow \infty$,$x^n \rightarrow 0$ only when $x \in (0, 1)$. If $x > 1$,then $x^n \rightarrow \infty$,and if $x = 1$,$\log _e 1 = 0$,but the limit behavior for $x^n$ is generally considered for $x \in (0, 1)$ to ensure the product vanishes. Thus,for the limit to be $0$,we must have $x \in (0, 1)$. Since the base of the logarithm $\log _x 12$ is $x \in (0, 1)$ and the argument $12 > 1$,the value of $\log _x 12$ must be negative.

If $\lim _{n \rightarrow \infty} x^n \log _e x=0$,then $\log _x 12=$

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