Question

Gabarito; -1/2
PRECISO DO CÁLCULO
$L = \lim _ { x \rightarrow 1 } ( \frac { x ^ { 3 } + x ^ { 2 } - 2 x } { x ^ { 2 } - 1 } )$
Obrigado.​

Answer 1

Resposta: 3/2

Explicação passo a passo:

$\lim_{x\to 1} \eft( \frac{x^{3}+x^2-2x}{x^2-1} )\right = \lim_{x\to 1} \eft( \frac{(x-1)(x^2+2x)}{(x+1)(x-1)} )\right = \lim_{x\to 1} \eft( \frac{x^2+2x}{x+1} )\right = \frac{\lim_{x\to 1}(x^2+2x)}{\lim_{x\to 1}(x+1)} = \frac{3}{2}$

Tem certeza desse gabarito?

Answer 2

Follow the steps.

Substituting x = 1 yields:

L = (1³ + 1² - 2 • 2)/(1² - 1) = 0/0.

Thus, applying L'Hopital's Rule,

L = lim(x → 1) (x³ + x² - 2x)'/(x² - 1)'

= lim(x → 1) (3x² + 2x - 2)/2x

= 1/2 • lim(x → 1) (3x + 1 - 1/x)

= 1/2 • (lim(x → 1) 3x - lim(x → 1) 1/x +1)

= 1/2(3 - 1 + 1)

= 1/2 • 3

= 3/2

Therefore,

L = lim(x → 1) (x³ + x² - 2x)/(x² - 1) = 3/2.

Yes I corrected my answer.

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