Question

AYUDA POR FAVOR!! con estos 5 problemas de limites en calculo!!!

Answer 1

$\lim_{x \to \infty} (\frac{x^{2}-4}{3x^{2}+x+2})=\lim_{x \to \infty} \dfrac{\frac{x^{2}}{x^{2}}-\frac{4}{x^{2} } }{\frac{3x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{2}{x^{2}}}= \lim_{x \to \infty} \frac{1-0}{3+0+0}=\frac{1}{3}$

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$\lim_{x \to -\infty} (\frac{2x}{x^{2}+4})= \lim_{x \to -\infty}(\dfrac{\frac{2x}{x^{2}}}{\frac{x^{2}}{x^{2} }+\frac{4}{x^{2}}})=\frac{0}{1+0}=0$

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$\lim_{x \to -\infty} \frac{\sqrt{4x^{2}+1}}{2x-3}=\lim_{x \to -\infty} \dfrac{\sqrt{x^{2}(4+\frac{1}{x^{2}})}}{2x-3}$

Aquí tener en cuenta que como x→-∞ entonces √x²=-x

$\lim_{x \to -\infty} \dfrac{\sqrt{x^{2}(4+\frac{1}{x^{2}})}}{2x-3}=\lim_{x \to -\infty} \dfrac{-x\sqrt{4+\frac{1}{x^{2}}}}{2x-3}=\lim_{x \to -\infty} \frac{\left(\dfrac{-x\sqrt{4+\frac{1}{x^{2}}}}{x}\right)}{\left(\dfrac{2x-3}{x}\right)}$

Después de simplificar:

$=\lim_{x \to -\infty} \dfrac{-\sqrt{4-\frac{1}{x^{2}}}}{2-\frac{3}{x}}=-\dfrac{\lim_{x \to -\infty \sqrt{4-\frac{1}{x^{2}}}}}{\lim_{x \to -\infty} (2-\frac{3}{x})}=-\frac{\sqrt{4-0}}{2-0}=-\frac{\sqrt{4}}{2}=-\frac{2}{2}=-1$

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$\lim_{x \to -\infty} \frac{3x^{3}+4x^{2}-5x+2}{6x^{2}-2x+3}=\lim_{x \to -\infty} \dfrac{\frac{3x^{3}}{x^{2}}+\frac{4x^{2}}{x^{2}}-\frac{5x}{x^{2}}+\frac{2}{x^{2}}}{\frac{6x^{2}}{x^{2}}-\frac{2x}{x^{2}}+\frac{3}{x^{2}}}\\=\lim_{x \to -\infty}\dfrac{3x+4-\frac{5}{x}+\frac{2}{x^{2}}}{6+\frac{2}{x}+\frac{3}{x^{2}}} =\dfrac{\lim_{x \to -\infty} 3x+4-\frac{5}{x}+\frac{2}{x^{2}}}{\lim_{x \to -\infty} 6+\frac{2}{x}+\frac{3}{x^{2}}}\\=\dfrac{-\infty+4-0+0}{6+0+0}=\frac{-\infty}{6}=-\infty$