lim_(x→1) ((((x^3 +7x^2 +10x+9))^(1/3) −(√(6x+3)))/((x−1)^2 ))


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Question Number 186065 by TUN last updated on 31/Jan/23

$\lim_{x \to 1} \frac{\sqrt[3]{x^{3} + 7 x^{2} + 10 x + 9} - \sqrt{6 x + 3}}{{(x - 1)}^{2}}$
Commented by Frix last updated on 01/Feb/23

$Use 2 times l^{'} H \hat{o p i t a l} \Rightarrow \frac{11}{54}$
Commented by TUN last updated on 01/Feb/23

$w i t h o u t L^{'} H o p i t a l$
	Answered by cortano1 last updated on 01/Feb/23
	$L_1 =lim_(x→1) ((((x^3 +7x^2 +10x+9))^(1/3) −(x+2))/((x−1)^2 )) = lim_(x→1) (((x−1)^2 )/((x−1)^2 {(((x^3 +7x^2 +10x+9)^2 ))^(1/3) +(x+2)((x^3 +7x^2 +10x+9))^(1/3) +(x+2)^2 })) = (1/(9+9+9))=(1/(27)) L_2 =lim_(x→1) ((x+2−(√(6x+3)))/((x−1)^2 )) = lim_(x→1) (((x−1)^2 )/((x−1)^2 {x+2+(√(6x+3))}=)) = (1/(3+3)) = (1/(6 )) ∴ L=L_1 +L_2 =(1/(27)) +(1/6)=((11)/(54))$
	$L_{1} = \lim_{x \to 1} \frac{\sqrt[3]{x^{3} + 7 x^{2} + 10 x + 9} - (x + 2)}{{(x - 1)}^{2}}$ $= \lim_{x \to 1} \frac{{(x - 1)}^{2}}{{(x - 1)}^{2} {\sqrt[3]{{(x^{3} + 7 x^{2} + 10 x + 9)}^{2}} + (x + 2) \sqrt[3]{x^{3} + 7 x^{2} + 10 x + 9} + {(x + 2)}^{2}}}$ $= \frac{1}{9 + 9 + 9} = \frac{1}{27}$ $L_{2} = \lim_{x \to 1} \frac{x + 2 - \sqrt{6 x + 3}}{{(x - 1)}^{2}}$ $= \lim_{x \to 1} \frac{{(x - 1)}^{2}}{{(x - 1)}^{2} {x + 2 + \sqrt{6 x + 3}} =}$ $= \frac{1}{3 + 3} = \frac{1}{6}$ $∴ L = L_{1} + L_{2} = \frac{1}{27} + \frac{1}{6} = \frac{11}{54}$
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