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![[tex]\frac{x}{\sqrt{x^2 - x - 1}} = \frac{x}{\sqrt{x^2}\sqrt{\displaystyle{1 - \frac{1}{x} - \frac{1}{x^2}}}} = \frac{x}{|x|} \cdot \frac{1}{\sqrt{\displaystyle{1 - \frac{1}{x} - \frac{1}{x^2}}}}[/tex]](images/latex/b0fd160fa997bfb05ffbad6f22c01856c32d31db_0.png "[tex]\frac{x}{\sqrt{x^2 - x - 1}} = \frac{x}{\sqrt{x^2}\sqrt{\displaystyle{1 - \frac{1}{x} - \frac{1}{x^2}}}} = \frac{x}{|x|} \cdot \frac{1}{\sqrt{\displaystyle{1 - \frac{1}{x} - \frac{1}{x^2}}}}[/tex]")
![[tex]\frac{x}{|x|}[/tex]](images/latex/32eaa2f77ed07b8022fa53ac65ef2c4f3995eaf6_0.png "[tex]\frac{x}{|x|}[/tex]")
para ![[tex]x \to -\infty[/tex]](images/latex/5e991aea98f95fce0aee96cea54a4243292f503a_0.png "[tex]x \to -\infty[/tex]")
?
![[tex]\frac{1-\sqrt{x}}{2-\sqrt{x+3}} = \frac{2+\sqrt{x+3}}{1+\sqrt{x}}\cdot \frac{1-x}{4 - (x + 3)}[/tex]](images/latex/1ab3872259785fc1dae8194b393cd1237e0cf4a4_0.png "[tex]\frac{1-\sqrt{x}}{2-\sqrt{x+3}} = \frac{2+\sqrt{x+3}}{1+\sqrt{x}}\cdot \frac{1-x}{4 - (x + 3)}[/tex]")
![[tex]x \to 1[/tex]](images/latex/efff73abf92757f763142664d5759a5051caed60_0.png "[tex]x \to 1[/tex]")
?
![[tex]\frac{4\sen x + \sqrt[3]{x^5}}{5x} = \frac{4}{5} \cdot \frac{\sen x}{x} + \frac{1}{5}\sqrt[3]{x^2}[/tex]](images/latex/46ac081e532d04ea18731eaddecfdc42a8317ba1_0.png "[tex]\frac{4\sen x + \sqrt[3]{x^5}}{5x} = \frac{4}{5} \cdot \frac{\sen x}{x} + \frac{1}{5}\sqrt[3]{x^2}[/tex]")
![[tex]\frac{\sen x}{x}[/tex]](images/latex/f0cbb7f1a82140c6b6cc239a16a7bfb1a6b02d9a_0.png "[tex]\frac{\sen x}{x}[/tex]")
para ![[tex]x \to 0[/tex]](images/latex/78672b47d1e994252af4d64692009f0fbdedcd5d_0.png "[tex]x \to 0[/tex]")
?
![[tex]\frac{1 - \cos x}{\sen^2{3x}} = \frac{1 - \cos x}{\sen^2{3x}}\cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{\sen^2{3x}} \cdot \frac{1}{1 + \cos x} =[/tex]](images/latex/4e506418df69e4fafb4ee6f68f1071bd91d0bf74_0.png "[tex]\frac{1 - \cos x}{\sen^2{3x}} = \frac{1 - \cos x}{\sen^2{3x}}\cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{\sen^2{3x}} \cdot \frac{1}{1 + \cos x} =[/tex]")
![[tex]\left ( \frac{\sen x}{x} \right )^2 \cdot \frac{1}{9} \cdot \left ( \frac{3x}{\sen 3x} \right )^2 \cdot \frac{1}{1 + \cos x}[/tex]](images/latex/97aefc415162538ea14e5e9a3ea1abfbedbd38cc_0.png "[tex]\left ( \frac{\sen x}{x} \right )^2 \cdot \frac{1}{9} \cdot \left ( \frac{3x}{\sen 3x} \right )^2 \cdot \frac{1}{1 + \cos x}[/tex]")
![[tex]\frac{\sen ax}{ax}[/tex]](images/latex/16493be39e77c582ab52b2f4477373b9b8a239f3_0.png "[tex]\frac{\sen ax}{ax}[/tex]")
para ![[tex]\frac{2^x - 5^x}{3^x + 1} = \frac{\left( \frac{2}{5}\right )^x - 1}{\left( \frac{3}{5}\right )^x + \left( \frac{1}{5}\right )^x}[/tex]](images/latex/be8b2bde43f21830a3edfb098263e3553ed69787_0.png "[tex]\frac{2^x - 5^x}{3^x + 1} = \frac{\left( \frac{2}{5}\right )^x - 1}{\left( \frac{3}{5}\right )^x + \left( \frac{1}{5}\right )^x}[/tex]")
![[tex]\lim_{x \to +\infty}\left( \frac{1}{5}\right )^x = \lim_{x \to +\infty}\left ( \frac{2}{5}\right )^x = \lim_{x \to +\infty}\left ( \frac{3}{5}\right )^x = 0[/tex]](images/latex/4e9406a9016d7d7b1dda1a86950d00b5ff5bf93f_0.png "[tex]\lim_{x \to +\infty}\left( \frac{1}{5}\right )^x = \lim_{x \to +\infty}\left ( \frac{2}{5}\right )^x = \lim_{x \to +\infty}\left ( \frac{3}{5}\right )^x = 0[/tex]")
ya que ![[tex]\frac{1}{5}[/tex]](images/latex/9ad0de9dccb287021ea15f7541a4740bd9419386_0.png "[tex]\frac{1}{5}[/tex]")
, ![[tex]\frac{2}{5}[/tex]](images/latex/da1b1235726e149e2d568fd03db724d8c850ace3_0.png "[tex]\frac{2}{5}[/tex]")
y ![[tex]\frac{3}{5}[/tex]](images/latex/15a1ec0e5b888c5dfd185d9430efbebb4d20a19b_0.png "[tex]\frac{3}{5}[/tex]")
son menores que 1.
![[tex]x \to +\infty[/tex]](images/latex/1d5bcec55dae694f423ce69cc2d603cf442b6719_0.png "[tex]x \to +\infty[/tex]")
? ¿Del denominador? ¿Del cociente (notar que el denominador es positivo para todo x real)?
![[tex]\log_2 (\sqrt{x + 1}) - \log_2 x = \log_2 \left ( \frac{\sqrt{x + 1}}{x} \right ) = \log_2 \sqrt{\frac{1}{x} + \frac{1}{x^2}}[/tex]](images/latex/4f95cf2f9403ae3d91e815905634d09257244264_0.png "[tex]\log_2 (\sqrt{x + 1}) - \log_2 x = \log_2 \left ( \frac{\sqrt{x + 1}}{x} \right ) = \log_2 \sqrt{\frac{1}{x} + \frac{1}{x^2}}[/tex]")
![[tex]\lim_{x \to +\infty}\frac{1}{x} = \lim_{x \to +\infty}\frac{1}{x^2} = 0[/tex]](images/latex/b6a312e50c1842b89206edab7414d2a3cc1a9c23_0.png "[tex]\lim_{x \to +\infty}\frac{1}{x} = \lim_{x \to +\infty}\frac{1}{x^2} = 0[/tex]")
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