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Límite de la función/ -pi/2/ tan(5*x)/ log(x-pi/2)/tan(5*x)

Límite de la función log(x-pi/2)/tan(5*x)

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Solución

Ha introducido [src] 
      /   /    pi\\
      |log|x - --||
      |   \    2 /|
 lim  |-----------|
   pi \  tan(5*x) /
x->--+             
   2
lim⁡x→π2+(log⁡(x−π2)tan⁡(5x))\lim_{x \to \frac{\pi}{2}^+}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right) 
Limit(log(x - pi/2)/tan(5*x), x, pi/2)
Método de l'Hopital
Tenemos la indeterminación de tipo
0/0,

tal que el límite para el numerador es
lim⁡x→π2+1tan⁡(5x)=0\lim_{x \to \frac{\pi}{2}^+} \frac{1}{\tan{\left(5 x \right)}} = 0
y el límite para el denominador es
lim⁡x→π2+1log⁡(2x−π2)=0\lim_{x \to \frac{\pi}{2}^+} \frac{1}{\log{\left(\frac{2 x - \pi}{2} \right)}} = 0
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
lim⁡x→π2+(log⁡(x−π2)tan⁡(5x))\lim_{x \to \frac{\pi}{2}^+}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right)
=
Introducimos una pequeña modificación de la función bajo el signo del límite
lim⁡x→π2+(log⁡(2x−π2)tan⁡(5x))\lim_{x \to \frac{\pi}{2}^+}\left(\frac{\log{\left(\frac{2 x - \pi}{2} \right)}}{\tan{\left(5 x \right)}}\right)
=
lim⁡x→π2+(ddx1tan⁡(5x)ddx1log⁡(2x−π2))\lim_{x \to \frac{\pi}{2}^+}\left(\frac{\frac{d}{d x} \frac{1}{\tan{\left(5 x \right)}}}{\frac{d}{d x} \frac{1}{\log{\left(\frac{2 x - \pi}{2} \right)}}}\right)
=
lim⁡x→π2+(−(2x−π)(−5tan⁡2(5x)−5)log⁡(x−π2)22tan⁡2(5x))\lim_{x \to \frac{\pi}{2}^+}\left(- \frac{\left(2 x - \pi\right) \left(- 5 \tan^{2}{\left(5 x \right)} - 5\right) \log{\left(x - \frac{\pi}{2} \right)}^{2}}{2 \tan^{2}{\left(5 x \right)}}\right)
=
lim⁡x→π2+(ddx(−(2x−π)log⁡(x−π2)22tan⁡2(5x))ddx1−5tan⁡2(5x)−5)\lim_{x \to \frac{\pi}{2}^+}\left(\frac{\frac{d}{d x} \left(- \frac{\left(2 x - \pi\right) \log{\left(x - \frac{\pi}{2} \right)}^{2}}{2 \tan^{2}{\left(5 x \right)}}\right)}{\frac{d}{d x} \frac{1}{- 5 \tan^{2}{\left(5 x \right)} - 5}}\right)
=
lim⁡x→π2+(10xlog⁡(x−π2)2tan⁡(5x)+10xlog⁡(x−π2)2tan⁡3(5x)−2xlog⁡(x−π2)xtan⁡2(5x)−πtan⁡2(5x)2−5πlog⁡(x−π2)2tan⁡(5x)−log⁡(x−π2)2tan⁡2(5x)−5πlog⁡(x−π2)2tan⁡3(5x)+πlog⁡(x−π2)xtan⁡2(5x)−πtan⁡2(5x)250tan⁡3(5x)25tan⁡4(5x)+50tan⁡2(5x)+25+50tan⁡(5x)25tan⁡4(5x)+50tan⁡2(5x)+25)\lim_{x \to \frac{\pi}{2}^+}\left(\frac{\frac{10 x \log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan{\left(5 x \right)}} + \frac{10 x \log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan^{3}{\left(5 x \right)}} - \frac{2 x \log{\left(x - \frac{\pi}{2} \right)}}{x \tan^{2}{\left(5 x \right)} - \frac{\pi \tan^{2}{\left(5 x \right)}}{2}} - \frac{5 \pi \log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan{\left(5 x \right)}} - \frac{\log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan^{2}{\left(5 x \right)}} - \frac{5 \pi \log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan^{3}{\left(5 x \right)}} + \frac{\pi \log{\left(x - \frac{\pi}{2} \right)}}{x \tan^{2}{\left(5 x \right)} - \frac{\pi \tan^{2}{\left(5 x \right)}}{2}}}{\frac{50 \tan^{3}{\left(5 x \right)}}{25 \tan^{4}{\left(5 x \right)} + 50 \tan^{2}{\left(5 x \right)} + 25} + \frac{50 \tan{\left(5 x \right)}}{25 \tan^{4}{\left(5 x \right)} + 50 \tan^{2}{\left(5 x \right)} + 25}}\right)
=
lim⁡x→π2+(10xlog⁡(x−π2)2tan⁡(5x)+10xlog⁡(x−π2)2tan⁡3(5x)−2xlog⁡(x−π2)xtan⁡2(5x)−πtan⁡2(5x)2−5πlog⁡(x−π2)2tan⁡(5x)−log⁡(x−π2)2tan⁡2(5x)−5πlog⁡(x−π2)2tan⁡3(5x)+πlog⁡(x−π2)xtan⁡2(5x)−πtan⁡2(5x)250tan⁡3(5x)25tan⁡4(5x)+50tan⁡2(5x)+25+50tan⁡(5x)25tan⁡4(5x)+50tan⁡2(5x)+25)\lim_{x \to \frac{\pi}{2}^+}\left(\frac{\frac{10 x \log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan{\left(5 x \right)}} + \frac{10 x \log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan^{3}{\left(5 x \right)}} - \frac{2 x \log{\left(x - \frac{\pi}{2} \right)}}{x \tan^{2}{\left(5 x \right)} - \frac{\pi \tan^{2}{\left(5 x \right)}}{2}} - \frac{5 \pi \log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan{\left(5 x \right)}} - \frac{\log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan^{2}{\left(5 x \right)}} - \frac{5 \pi \log{\left(x - \frac{\pi}{2} \right)}^{2}}{\tan^{3}{\left(5 x \right)}} + \frac{\pi \log{\left(x - \frac{\pi}{2} \right)}}{x \tan^{2}{\left(5 x \right)} - \frac{\pi \tan^{2}{\left(5 x \right)}}{2}}}{\frac{50 \tan^{3}{\left(5 x \right)}}{25 \tan^{4}{\left(5 x \right)} + 50 \tan^{2}{\left(5 x \right)} + 25} + \frac{50 \tan{\left(5 x \right)}}{25 \tan^{4}{\left(5 x \right)} + 50 \tan^{2}{\left(5 x \right)} + 25}}\right)
=
00
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)
Gráfica
-3.0-2.5-2.0-1.5-1.0-0.50.00.51.01.52.02.53.0-10000000000000001000000000000000
Trazar el gráfico
A la izquierda y a la derecha [src] 
      /   /    pi\\
      |log|x - --||
      |   \    2 /|
 lim  |-----------|
   pi \  tan(5*x) /
x->--+             
   2
lim⁡x→π2+(log⁡(x−π2)tan⁡(5x))\lim_{x \to \frac{\pi}{2}^+}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right) 
0
00 
= 0.00966883345446983
      /   /    pi\\
      |log|x - --||
      |   \    2 /|
 lim  |-----------|
   pi \  tan(5*x) /
x->---             
   2
lim⁡x→π2−(log⁡(x−π2)tan⁡(5x))\lim_{x \to \frac{\pi}{2}^-}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right) 
0
00 
= (-0.00974778342223482 + 0.00401661837640065j)
= (-0.00974778342223482 + 0.00401661837640065j)
Respuesta rápida [src] 
0
00
Abrir y simplificar
Otros límites con x→0, -oo, +oo, 1
lim⁡x→π2−(log⁡(x−π2)tan⁡(5x))=0\lim_{x \to \frac{\pi}{2}^-}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right) = 0
Más detalles con x→pi/2 a la izquierda
lim⁡x→π2+(log⁡(x−π2)tan⁡(5x))=0\lim_{x \to \frac{\pi}{2}^+}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right) = 0
lim⁡x→∞(log⁡(x−π2)tan⁡(5x))\lim_{x \to \infty}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right)
Más detalles con x→oo
lim⁡x→0−(log⁡(x−π2)tan⁡(5x))=−∞sign⁡(−log⁡(2)+log⁡(π)+iπ)\lim_{x \to 0^-}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right) = - \infty \operatorname{sign}{\left(- \log{\left(2 \right)} + \log{\left(\pi \right)} + i \pi \right)}
Más detalles con x→0 a la izquierda
lim⁡x→0+(log⁡(x−π2)tan⁡(5x))=∞sign⁡(−log⁡(2)+log⁡(π)+iπ)\lim_{x \to 0^+}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right) = \infty \operatorname{sign}{\left(- \log{\left(2 \right)} + \log{\left(\pi \right)} + i \pi \right)}
Más detalles con x→0 a la derecha
lim⁡x→1−(log⁡(x−π2)tan⁡(5x))=log⁡(−1+π2)+iπtan⁡(5)\lim_{x \to 1^-}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right) = \frac{\log{\left(-1 + \frac{\pi}{2} \right)} + i \pi}{\tan{\left(5 \right)}}
Más detalles con x→1 a la izquierda
lim⁡x→1+(log⁡(x−π2)tan⁡(5x))=log⁡(−1+π2)+iπtan⁡(5)\lim_{x \to 1^+}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right) = \frac{\log{\left(-1 + \frac{\pi}{2} \right)} + i \pi}{\tan{\left(5 \right)}}
Más detalles con x→1 a la derecha
lim⁡x→−∞(log⁡(x−π2)tan⁡(5x))\lim_{x \to -\infty}\left(\frac{\log{\left(x - \frac{\pi}{2} \right)}}{\tan{\left(5 x \right)}}\right)
Más detalles con x→-oo
Respuesta numérica [src] 
0.00966883345446983
0.00966883345446983
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