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inverse of 20((x-3)(x+4))/((x-3)^2(x-5))

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

☐²

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ddx

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

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lim_{x→ 0}(1−sec²(7x)(3x)² )

Solution

−499

Decimal

−5.44444…

Mixed Number

−549

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Solution steps

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lim_{x→ 0}(1−sec²(7x)(3x)² )

Reeecreva usando identidades trigonométricas

=lim_{x→ 0}(−tan²(7x)9x² )

lim_x→a[c·f(x)]=c·lim_x→af(x)

=−19 · lim_{x→ 0}(tan²(7x)x² )

Simplificar tan²(7x)x² : (tan(7x)x )²

=−19 · lim_{x→ 0}((tan(7x)x )²)

lim_x→a[f(x)]^b=[lim_x→af(x)]^b
Com exceção da forma indeterminada

=−19 (lim_{x→ 0}(tan(7x)x ))²

Aplicar a regra de L'Hopital: lim_{x→ 0}(sec²(7x)· 71 )

=−19 (lim_{x→ 0}(sec²(7x)· 71 ))²

Inserir o valor

=−19 (sec²(7· 0)· 71 )²

Simplificar −19 (sec²(7· 0)· 71 )²: −499

=−499

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

inverse of 20((x-3)(x+4))/((x-3)^2(x-5))

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