Question

Lim que tende ao infinito x2-16/x5+4

Answer 1

$\lim_{x \to \infty} \frac{x^2 - 16}{x^5 + 4} \\ \lim_{x \to \infty} \frac{x^2.[1 - \frac{16}{x^2}] }{x^2.[ x^3 + \frac{4}{x^2}] }$

$lim_{x \to \infty} \frac{ 1 }{ x^3 } = 0$

Answer 2

Vamos lá:

$\lim _{x\to \infty \:}\left(\frac{x^2-16}{x^5+4}\right)\\\lim _{x\to \infty \:}\left(\frac{\frac{x^2}{x^5}-\frac{16}{x^5}}{\frac{x^5}{x^5}+\frac{4}{x^5}}\right)\\\lim \:_{x\to \:\infty \:\:}\left(\frac{\frac{1}{x^3}-\frac{16}{x^5}}{1+\frac{4}{x^5}}\right)=\frac{0-0}{1+0}=\frac{0}{1}=0$
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Método pratico de resolver: considere apenas "x" de maior exponente tanto no denominar quanto numerador:

$\lim \:_{x\to \:\infty }\left(\frac{x^2-16}{x^5+4}\right)=\lim \:_{x\to \:\infty }\left(\frac{x^2}{x^5}\right)=\lim \:\:_{x\to \:\:\infty }\left(\frac{1}{x^3}\right)=0$
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Espero ter ajudado!