inverse of 4^{-2x+2}

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
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simplify solve for inverse tangent line

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lim_{x→ ∞}(x⁴−x²+2x²(x−1) )

Solution

∞

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lim_{x→ ∞}(x⁴−x²+2x²(x−1) )

Apply the following algebraic property: a+b=a(1+ba )
x⁴−x²+2x²(x−1) =x⁴(1−1x² +2x⁴ )x²(x(1−1x ))

=lim_{x→ ∞}(x⁴(1−1x² +2x⁴ )x²x(1−1x ) )

Simplify x⁴(1−1x² +2x⁴ )x²(x(1−1x )) : x(2x⁴ −1x² +1)1−1x

=lim_{x→ ∞}(x(2x⁴ −1x² +1)1−1x )

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→ ∞}(x(2x⁴ −1x² +1))lim_{x→ ∞}(1−1x )

lim_{x→ ∞}(x(2x⁴ −1x² +1))=∞

lim_{x→ ∞}(1−1x )=1

=∞ 1

Apply Infinity Property: ∞c =∞

=∞

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

inverse of 4^{-2x+2}

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