lim_(x→2) (((x^4 −4x^3 +5x^2 −4x+4))^(1/(4 )) /( (√(x^2 −3x+2)))) = ?


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Question Number 115551 by bemath last updated on 26/Sep/20

$\lim_{x \to 2} \frac{\sqrt[4]{x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4}}{\sqrt{x^{2} - 3 x + 2}} = ?$
	Answered by TANMAY PANACEA last updated on 26/Sep/20
	$x^2 −3x+2=(x−1)(x−2) x^4 −4x^3 +5x^2 −4x+4 =x^4 −2x^3 −2x^3 +4x^2 +x^2 −2x−2x+4 =x^3 (x−2) −2x^2 (x−2) +x(x−2) −2 (x−2) (x−2)(x^3 −2x^2 +x−2) =(x−2){x^2 (x−2)+1(x−2)} =(x−2)^2 (x^2 +1) so N_r =(x−2)^(1/2) ×(x^2 +1)^(1/4) D_r =(x−2)^(1/2) ×(x−1)^(1/2) lim_(x→2) (((x−2)^(1/2) ×(x^2 +1)^(1/4) )/((x−2)^(1/2) ×(x−1)^(1/2) ))=(((5)^(1/4) )/1)$
	$x^{2} - 3 x + 2 = (x - 1) (x - 2)$ $x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4$ $= x^{4} - 2 x^{3} - 2 x^{3} + 4 x^{2} + x^{2} - 2 x - 2 x + 4$ $= x^{3} (x - 2) - 2 x^{2} (x - 2) + x (x - 2) - 2 (x - 2)$ $(x - 2) (x^{3} - 2 x^{2} + x - 2)$ $= (x - 2) {x^{2} (x - 2) + 1 (x - 2)}$ $= {(x - 2)}^{2} (x^{2} + 1)$ $s o N_{r} = {(x - 2)}^{\frac{1}{2}} \times {(x^{2} + 1)}^{\frac{1}{4}}$ $D_{r} = {(x - 2)}^{\frac{1}{2}} \times {(x - 1)}^{\frac{1}{2}}$ $\lim_{x \to 2} \frac{{(x - 2)}^{\frac{1}{2}} \times {(x^{2} + 1)}^{\frac{1}{4}}}{{(x - 2)}^{\frac{1}{2}} \times {(x - 1)}^{\frac{1}{2}}} = \frac{{(5)}^{\frac{1}{4}}}{1}$
	Commented by bemath last updated on 26/Sep/20

	$y e s s$
	Answered by bemath last updated on 26/Sep/20

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