Consideremos el siguiente límite
![\lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{tg(x)} = (+\infty)^0 \lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{tg(x)} = (+\infty)^0](local/cache-vignettes/L169xH44/7461b5c377b49c2d4f41691f5887af1f-d9ad5.png?1688141968)
Función 1/x
LA función 1/x tiendo a +infinito cuando x se acerca a 0 por la derecha
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Llamamos "A" al límite y tomamos logaritmos
![\lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{tg(x)} = A \lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{tg(x)} = A](local/cache-vignettes/L132xH44/aad4d8ee97982cccf02d4319a56c0756-b0c34.png?1688141968)
![ln \left[ \lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{tg(x)} \right] = ln(A) ln \left[ \lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{tg(x)} \right] = ln(A)](local/cache-vignettes/L196xH50/b1c7da9a109af8178eff4de676da0e58-d552b.png?1688141968)
![\lim_{x \rightarrow 0^+} \left[ ln \left( \frac{1}{x} \right)^{tg(x)} \right] = ln(A) \lim_{x \rightarrow 0^+} \left[ ln \left( \frac{1}{x} \right)^{tg(x)} \right] = ln(A)](local/cache-vignettes/L196xH50/28c5cd2c0fc3ca905fde5f3bae46e56d-c03d1.png?1688141968)
![\lim_{x \rightarrow 0^+} \left[ (tg(x)) \cdot ln \left( \frac{1}{x} \right)\right] = ln(A) \lim_{x \rightarrow 0^+} \left[ (tg(x)) \cdot ln \left( \frac{1}{x} \right)\right] = ln(A)](local/cache-vignettes/L229xH40/5c671712211b14e42592928ba2f0cb31-e967e.png?1688141968)
Nos centramos en la expresión de la izquierda del signo igual
![\lim_{x \rightarrow 0^+} \left[ (tg(x)) \cdot ln \left( \frac{1}{x} \right)\right] = 0 \cdot (+\infty) \lim_{x \rightarrow 0^+} \left[ (tg(x)) \cdot ln \left( \frac{1}{x} \right)\right] = 0 \cdot (+\infty)](local/cache-vignettes/L251xH40/663c5f42dcde41dd55b4e824c43e2311-8d116.png?1688141968)
Función ln(1/x)
función logaritmo neperiano de 1/x
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Se ha transformado en una indeterminación del tipo
que vimos cómo resolver en indeterminación 0 · infinito
![\lim_{x \rightarrow 0^+} \left[ (tg(x)) \cdot ln \left( \frac{1}{x} \right)\right] = \lim_{x \rightarrow 0^+} \frac{ln \left( \frac{1}{x} \right)}{1/tg(x)} = \frac{+\infty}{+\infty} \lim_{x \rightarrow 0^+} \left[ (tg(x)) \cdot ln \left( \frac{1}{x} \right)\right] = \lim_{x \rightarrow 0^+} \frac{ln \left( \frac{1}{x} \right)}{1/tg(x)} = \frac{+\infty}{+\infty}](local/cache-vignettes/L335xH43/652e27014ec7dcce6209f74a0d5a5481-7c449.png?1688141968)
Ahora podemos aplicar L’Hôpital
![\lim_{x \rightarrow 0^+}\frac{ln \left( \frac{1}{x} \right)}{1/tg(x)} \stackrel{l \:^\prime H}{=} \lim_{x \rightarrow 0^+} \frac{-1/x^2}{-1/sen^2(x)} = \lim_{x \rightarrow 0^+} \frac{sen^2(x)}{x} = \frac{0}{0} \lim_{x \rightarrow 0^+}\frac{ln \left( \frac{1}{x} \right)}{1/tg(x)} \stackrel{l \:^\prime H}{=} \lim_{x \rightarrow 0^+} \frac{-1/x^2}{-1/sen^2(x)} = \lim_{x \rightarrow 0^+} \frac{sen^2(x)}{x} = \frac{0}{0}](local/cache-vignettes/L389xH43/72bc823fa7cfabf59a4285b8e0c12bee-6323e.png?1688141968)
Volvemos a aplicar L’Hôpital
![\lim_{x \rightarrow 0^+} \frac{sen^2(x)}{x} \stackrel{l \:^\prime H}{=} \frac{2 \cdot sen (x) ─\cdot cos(x)}{1} = \frac{0}{1} = 0 \lim_{x \rightarrow 0^+} \frac{sen^2(x)}{x} \stackrel{l \:^\prime H}{=} \frac{2 \cdot sen (x) ─\cdot cos(x)}{1} = \frac{0}{1} = 0](local/cache-vignettes/L312xH37/848831711d285862c2341d2195ccc2d6-e2ab6.png?1688141968)
Entonces, como teníamos que:
![\lim_{x \rightarrow 0^+} \left[ (tg(x)) \cdot ln \left( \frac{1}{x} \right)\right] = ln(A) \lim_{x \rightarrow 0^+} \left[ (tg(x)) \cdot ln \left( \frac{1}{x} \right)\right] = ln(A)](local/cache-vignettes/L229xH40/5c671712211b14e42592928ba2f0cb31-e967e.png?1688141968)
![0 = ln(A) \longrightarrow \fbox {A=1} 0 = ln(A) \longrightarrow \fbox {A=1}](local/cache-vignettes/L149xH22/ae105cb8f785ef4628828ee59c3c20e7-41301.png?1688141968)
Por tanto
![\lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{tg(x)} = 1 \lim_{x \rightarrow 0^+} \left( \frac{1}{x} \right)^{tg(x)} = 1](local/cache-vignettes/L127xH44/6670064d7a56b798f5b25ded82007a38-bf2ad.png?1688141968)