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degree (2x+3)^3-4x^3

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lim_{x→ ∞}(x−4x²−5x+6 )

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lim_{x→ ∞}(x−4x²−5x+6 )

Divide by highest denominator power: 1x −4x² 1−5x +6x²

=lim_{x→ ∞}(1x −4x² 1−5x +6x² )

lim_x→a[f(x)g(x) ]=lim_x→af(x)lim_x→ag(x) , lim_x→ag(x)≠0
With the exception of indeterminate form

=lim_{x→ ∞}(1x −4x² )lim_{x→ ∞}(1−5x +6x² )

lim_{x→ ∞}(1x −4x² )=0

lim_{x→ ∞}(1−5x +6x² )=1

=01

Apply rule 0a =0, a≠ 0

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

degree (2x+3)^3-4x^3

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