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inversa -4^{x+1}

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lim_{x→ ∞}(2x²x²−1 )

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lim_{x→ ∞}(2x²x²−1 )

lim_x→a[c·f(x)]=c·lim_x→af(x)

=2· lim_{x→ ∞}(x²x²−1 )

Dividir pelo denominador de maior potência: 11−1x²

=2· lim_{x→ ∞}(11−1x² )

lim_x→a[f(x)g(x) ]=lim_x→af(x)lim_x→ag(x) , lim_x→ag(x)≠0
Com exceção da forma indeterminada

=2· lim_{x→ ∞}(1)lim_{x→ ∞}(1−1x² )

lim_{x→ ∞}(1)=1

lim_{x→ ∞}(1−1x² )=1

=2· 11

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

inversa -4^{x+1}

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