calculste lim_(x→0) ((arctan(x^2 +x+1)−arctan(x+1))/x^3 )


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Question Number 123675 by mathmax by abdo last updated on 27/Nov/20

$calculste \lim_{x \to 0} \frac{\arctan (x^{2} + x + 1) - \arctan (x + 1)}{x^{3}}$
	Answered by benjo_mathlover last updated on 27/Nov/20
	$lim_(x→0) ((((2x+1)/(1+(x^2 +x+1)^2 )) −(1/(1+(x+1)^2 )))/(3x^2 )) = lim_(x→0) (((2x+1)((1+x^2 +2x+1)−(1+(x^2 +x+1)^2 ))/(3x^2 {(1+(x^2 +x+1)^2 )(1+(1+x)^2 }))= (1/4).lim_(x→0) ((2x^3 +5x^2 +6x+2−x^4 −2x^3 −3x^2 −2x−2)/(3x^2 ))= (1/4). lim_(x→0) ((−x^4 +2x^2 +4x)/(3x^2 )) = (1/4).lim_(x→0) ((−x^3 +2x+4)/(3x)) = ∞$
	$\lim_{x \to 0} \frac{\frac{2 x + 1}{1 + {(x^{2} + x + 1)}^{2}} - \frac{1}{1 + {(x + 1)}^{2}}}{3 x^{2}} =$ $\lim_{x \to 0} \frac{(2 x + 1) ((1 + x^{2} + 2 x + 1) - (1 + {(x^{2} + x + 1)}^{2})}{3 x^{2} {(1 + {(x^{2} + x + 1)}^{2}) (1 + {(1 + x)}^{2}}} =$ $\frac{1}{4} . \lim_{x \to 0} \frac{2 x^{3} + 5 x^{2} + 6 x + 2 - x^{4} - 2 x^{3} - 3 x^{2} - 2 x - 2}{3 x^{2}} =$ $\frac{1}{4} . \lim_{x \to 0} \frac{- x^{4} + 2 x^{2} + 4 x}{3 x^{2}} =$ $\frac{1}{4} . \lim_{x \to 0} \frac{- x^{3} + 2 x + 4}{3 x} = \infty$
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