lim _{x rightarrow infty}left(frac{3 x^2-2 x+ | vedclass

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(C) The given limit is of the form $1^\infty$. We use the formula $\lim _{x \rightarrow \infty} (f(x))^{g(x)} = e^{\lim _{x \rightarrow \infty} (f(x)-

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$e^{-3}$

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(C) The given limit is of the form $1^\infty$. We use the formula $\lim _{x \rightarrow \infty} (f(x))^{g(x)} = e^{\lim _{x \rightarrow \infty} (f(x)-1)g(x)}$. Here,$f(x) = \frac{3x^2-2x+3}{3x^2+x-2}$ and $g(x) = 3x-2$. $(f(x)-1) = \frac{3x^2-2x+3 - (3x^2+x-2)}{3x^2+x-2} = \frac{-3x+5}{3x^2+x-2}$. Now,$\lim _{x \rightarrow \infty} (f(x)-1)g(x) = \lim _{x \rightarrow \infty} \left(\frac{-3x+5}{3x^2+x-2}\right)(3x-2)$. $= \lim _{x \rightarrow \infty} \frac{-9x^2+6x+15x-10}{3x^2+x-2} = \lim _{x \rightarrow \infty} \frac{-9x^2+21x-10}{3x^2+x-2}$. Dividing the numerator and denominator by $x^2$,we get $\frac{-9}{3} = -3$. Therefore,the limit is $e^{-3}$.

$\lim _{x \rightarrow \infty}\left(\frac{3 x^2-2 x+3}{3 x^2+x-2}\right)^{3 x-2} = $

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