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

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(B) Let $L = \lim _{x \rightarrow \infty} \left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}}$.As $x \rightarrow \infty$,$|x| = x$.So,$

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$2 \sqrt{2}$

Vedclass · Answer

(B) Let $L = \lim _{x ightarrow \infty} \left(\frac{2 x^2+3 x+4}{x^2-3 x+5}ight)^{\frac{3|x|+1}{2|x|-1}}$.As $x ightarrow \infty$,$|x| = x$.So,$L = \lim _{x}$ ${ightarrow \infty} \left(\frac{2 + \frac{3}{x} + \frac{4}{x^2}}{1 - \frac{3}{x} + \frac{5}{x^2}}ight)^{\frac{3 + \frac{1}{x}}{2 - \frac{1}{x}}}$.Evaluating the limit as $x ightarrow \infty$,the terms $\frac{1}{x}$ and $\frac{1}{x^2}$ approach $0$.Thus,$L = \left(\frac{2+0+0}{1-0+0}ight)^{\frac{3+0}{2-0}} = (2)^{\frac{3}{2}}$.$L = 2^{\frac{3}{2}} = 2^1 	imes 2^{\frac{1}{2}} = 2 \sqrt{2}$.

$\lim _{x \rightarrow \infty}\left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}} = $

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