lim _{x rightarrow 2}left(frac{5 x-8}{8-3 x}r | vedclass

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(D) The limit is of the form $1^\infty$. We use the formula $\lim_{x 	o a} f(x)^{g(x)} = e^{\lim_{x 	o a} (f(x)-1)g(x)}$.$\lim _{x ightarrow 2}\le

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$e^6$

Vedclass · Answer

(D) The limit is of the form $1^\infty$. We use the formula $\lim_{x 	o a} f(x)^{g(x)} = e^{\lim_{x 	o a} (f(x)-1)g(x)}$.$\lim _{x ightarrow 2}\left(\frac{5 x-8}{8-3 x}ight)^{\frac{3}{2 x-4}} = e^{\lim _{x ightarrow 2}\left(\frac{5 x-8}{8-3 x}-1ight) 	imes \frac{3}{2 x-4}}$$= e^{\lim _{x ightarrow 2}\left(\frac{5 x-8 - (8-3 x)}{8-3 x}ight) 	imes \frac{3}{2(x-2)}}$$= e^{\lim _{x ightarrow 2}\left(\frac{8x-16}{8-3 x}ight) 	imes \frac{3}{2(x-2)}}$$= e^{\lim _{x ightarrow 2}\left(\frac{8(x-2)}{8-3 x}ight) 	imes \frac{3}{2(x-2)}}$$= e^{\lim _{x ightarrow 2} \frac{24}{2(8-3 x)}}$$= e^{\frac{12}{8-3(2)}} = e^{\frac{12}{2}} = e^6$

$\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}} = $

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