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Guest2
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I'm trying to find $\displaystyle \lim_{x \to 20^{+}}\frac{5x^3+1}{20x^3-8000x}$
$\displaystyle \lim_{x \to 20^{+}}\frac{5x^3+1}{20x^3-8000x} =\lim_{x \to 20^{+}}\frac{5+1/x^3}{20-8000/x^2} = \frac{5+\lim_{x \to 20^{+}}1/x^3}{20-\lim_{x \to 20^{+}}8000/x^2} = \frac{5+\frac{1}{8000}}{20-\frac{8000}{400}} = \infty. $
I'm not sure because it seems I have a zero dominator throughout.
$\displaystyle \lim_{x \to 20^{+}}\frac{5x^3+1}{20x^3-8000x} =\lim_{x \to 20^{+}}\frac{5+1/x^3}{20-8000/x^2} = \frac{5+\lim_{x \to 20^{+}}1/x^3}{20-\lim_{x \to 20^{+}}8000/x^2} = \frac{5+\frac{1}{8000}}{20-\frac{8000}{400}} = \infty. $
I'm not sure because it seems I have a zero dominator throughout.