inverse of laplace 1/(x^3)

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

☐²

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≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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lim_{x→ ∞}(3x²−x−25x²+4x+1 )

Solution

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0.6

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lim_{x→ ∞}(3x²−x−25x²+4x+1 )

Divide by highest denominator power: 3−1x −2x² 5+4x +1x²

=lim_{x→ ∞}(3−1x −2x² 5+4x +1x² )

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→ ∞}(3−1x −2x² )lim_{x→ ∞}(5+4x +1x² )

lim_{x→ ∞}(3−1x −2x² )=3

lim_{x→ ∞}(5+4x +1x² )=5

=35

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

inverse of laplace 1/(x^3)

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