Question

Olá pode resolver esses itens :
a) 5×2-4×+3 / ×2+2 , x tendendo ao infinito
b) ×2 +5 × +2 / 2×3 -7, x tendendo ao infinito
c) × -1 / raiz quadrada x2 +8 -3 , x tendendo a 1

Answer 1

Oi Debora :)

Segue: 

$\lim_{x \to \infty} \frac{5x^2-4x+3}{x^2+2} \ \ \ \ coloca \ x^2 \ em \ evide^ncia\\ \\ \lim_{x \to \infty} \frac{x^2(5- \frac{4}{x} + \frac{3}{x^2}) }{x^2(1+ \frac{2}{x^2} )} \ \ \ \ cancelando \ x^2 \\ \\ \lim_{x \to \infty} \frac{5- \frac{4}{x} + \frac{3}{x} }{1+ \frac{2}{x^2} } \ \ \ substituindo \ x \ por \ \infty \\ \\ \lim_{x \to \infty} \frac{5- \frac{4}{\infty} + \frac{3}{\infty^2} }{1+ \frac{2}{\infty^2} } \ \ \ \ sabemos \ que \ n/\infty = 0$

$lim_{x \to \infty} \frac{5- 0 + 0 }{1+ 0 } \\ \\ lim_{x \to \infty} \ \ 5$

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$\lim_{x \to \infty} \frac{x^2+5x+2}{2x^3-7} \\ \\ \lim_{x \to \infty} \frac{x^2(1+ \frac{5}{x} + \frac{2}{x^2}) }{x^2(2x- \frac{7}{x^2} )} \\ \\ \lim_{x \to \infty} \frac{1+ \frac{5}{\infty} + \frac{2}{\infty^2} }{2(\infty)- \frac{7}{\infty^2} } \\ \\ \lim_{x \to \infty} \frac{1+0 + 0 }{\infty- 0 } \\ \\ \lim_{x \to \infty} \frac{1}{\infty } \\ \\ \lim_{x \to \infty} \ 0$

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$\lim_{x \to1} \frac{x-1}{ \sqrt{x^2+8x-3} } \ \ \ So \ substituir \ x\ por \ 1 \\ \\ \lim_{x \to1} \frac{1-1}{ \sqrt{1^2+8.1-3} } \\ \\ \lim_{x \to1} \frac{0}{ \sqrt{6} } \\ \\ \lim_{x \to1} \ 0$