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Límite de (1-cos(5*x))/x^2
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Límite de x^(1/log(-1+e^x))
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Límite de (2*x/(1+2*x))^x
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Límite de (5+x)/(-6+3*x)
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Expresiones idénticas
- sinh(pi/n)/sinh(pi/(uno +n))
- seno hiperbólico de ( número pi dividir por n) dividir por seno hiperbólico de ( número pi dividir por (1 más n))
- seno hiperbólico de ( número pi dividir por n) dividir por seno hiperbólico de ( número pi dividir por (uno más n))
- sinhpi/n/sinhpi/1+n
- sinh(pi dividir por n) dividir por sinh(pi dividir por (1+n))
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Expresiones semejantes
- sinh(pi/n)/sinh(pi/(1-n))
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Expresiones con funciones
- Seno hiperbólico sinh
- sinh(a-x)/cosh(a*x)
- sinh(-1+2*x)^3/sin(4*x)
- sinh(2*z^2)/z^3
- sinh(3*x)/(3*x)
- sinh(x)/(-x+x*e^x)
- Número Pi pi
- Piecewise((-9/2+9*x/2+9*((-1+x)^2)^(1/3)/2,x<2),(2-x-1/(-2+x),x>2))
- pi*acot(a)
- pi*x*(-2+x)/tan(x)
- Piecewise((2+x,x<=1),(log(-1+x)/log(2),True))
- pi*(68-x)/64
- Seno hiperbólico sinh
- sinh(a-x)/cosh(a*x)
- sinh(-1+2*x)^3/sin(4*x)
- sinh(2*z^2)/z^3
- sinh(3*x)/(3*x)
- sinh(x)/(-x+x*e^x)
- Número Pi pi
- Piecewise((-9/2+9*x/2+9*((-1+x)^2)^(1/3)/2,x<2),(2-x-1/(-2+x),x>2))
- pi*acot(a)
- pi*x*(-2+x)/tan(x)
- Piecewise((2+x,x<=1),(log(-1+x)/log(2),True))
- pi*(68-x)/64
Límite de la función sinh(pi/n)/sinh(pi/(1+n))
Solución
Ha introducido
[src]
/ /pi\ \
| sinh|--| |
| \n / |
lim |-----------|
n->oo| / pi \|
|sinh|-----||
\ \1 + n//
$$\lim_{n \to \infty}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right)$$
Limit(sinh(pi/n)/sinh(pi/(1 + n)), n, oo, dir='-')
Método de l'Hopital
Tenemos la indeterminación de tipo
tal que el límite para el numerador es
$$\lim_{n \to \infty} \frac{1}{\sinh{\left(\frac{\pi}{n + 1} \right)}} = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty} \frac{1}{\sinh{\left(\frac{\pi}{n} \right)}} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{1}{\sinh{\left(\frac{\pi}{n + 1} \right)}}}{\frac{d}{d n} \frac{1}{\sinh{\left(\frac{\pi}{n} \right)}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)} \cosh{\left(\frac{\pi}{n + 1} \right)}}{\left(n + 1\right)^{2} \sinh^{2}{\left(\frac{\pi}{n + 1} \right)} \cosh{\left(\frac{\pi}{n} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{1}{\left(\frac{1}{\sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2}{n \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{1}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}}\right) \sinh^{2}{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{1}{\sinh^{2}{\left(\frac{\pi}{n + 1} \right)}}}{\frac{d}{d n} \left(\frac{1}{\sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2}{n \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{1}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}}\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{2 \pi \cosh{\left(\frac{\pi}{n + 1} \right)}}{\left(n + 1\right)^{2} \left(- \frac{2}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{2} \sinh^{3}{\left(\frac{\pi}{n} \right)}} - \frac{2}{n^{3} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{4 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{3} \sinh^{3}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{4} \sinh^{3}{\left(\frac{\pi}{n} \right)}}\right) \sinh^{3}{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{2 \pi}{\left(n + 1\right)^{2} \left(- \frac{2}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{2} \sinh^{3}{\left(\frac{\pi}{n} \right)}} - \frac{2}{n^{3} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{4 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{3} \sinh^{3}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{4} \sinh^{3}{\left(\frac{\pi}{n} \right)}}\right) \sinh^{3}{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{2 \pi}{\left(n + 1\right)^{2} \left(- \frac{2}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{2} \sinh^{3}{\left(\frac{\pi}{n} \right)}} - \frac{2}{n^{3} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{4 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{3} \sinh^{3}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{4} \sinh^{3}{\left(\frac{\pi}{n} \right)}}\right) \sinh^{3}{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$1$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)
oo/oo,
tal que el límite para el numerador es
$$\lim_{n \to \infty} \frac{1}{\sinh{\left(\frac{\pi}{n + 1} \right)}} = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty} \frac{1}{\sinh{\left(\frac{\pi}{n} \right)}} = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{1}{\sinh{\left(\frac{\pi}{n + 1} \right)}}}{\frac{d}{d n} \frac{1}{\sinh{\left(\frac{\pi}{n} \right)}}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)} \cosh{\left(\frac{\pi}{n + 1} \right)}}{\left(n + 1\right)^{2} \sinh^{2}{\left(\frac{\pi}{n + 1} \right)} \cosh{\left(\frac{\pi}{n} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{1}{\left(\frac{1}{\sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2}{n \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{1}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}}\right) \sinh^{2}{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{1}{\sinh^{2}{\left(\frac{\pi}{n + 1} \right)}}}{\frac{d}{d n} \left(\frac{1}{\sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2}{n \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{1}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}}\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{2 \pi \cosh{\left(\frac{\pi}{n + 1} \right)}}{\left(n + 1\right)^{2} \left(- \frac{2}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{2} \sinh^{3}{\left(\frac{\pi}{n} \right)}} - \frac{2}{n^{3} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{4 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{3} \sinh^{3}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{4} \sinh^{3}{\left(\frac{\pi}{n} \right)}}\right) \sinh^{3}{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{2 \pi}{\left(n + 1\right)^{2} \left(- \frac{2}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{2} \sinh^{3}{\left(\frac{\pi}{n} \right)}} - \frac{2}{n^{3} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{4 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{3} \sinh^{3}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{4} \sinh^{3}{\left(\frac{\pi}{n} \right)}}\right) \sinh^{3}{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{2 \pi}{\left(n + 1\right)^{2} \left(- \frac{2}{n^{2} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{2} \sinh^{3}{\left(\frac{\pi}{n} \right)}} - \frac{2}{n^{3} \sinh^{2}{\left(\frac{\pi}{n} \right)}} + \frac{4 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{3} \sinh^{3}{\left(\frac{\pi}{n} \right)}} + \frac{2 \pi \cosh{\left(\frac{\pi}{n} \right)}}{n^{4} \sinh^{3}{\left(\frac{\pi}{n} \right)}}\right) \sinh^{3}{\left(\frac{\pi}{n + 1} \right)}}\right)$$
=
$$1$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)
Gráfica
Otros límites con n→0, -oo, +oo, 1
$$\lim_{n \to \infty}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = -\infty$$
Más detalles con n→0 a la izquierda
$$\lim_{n \to 0^+}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = \infty$$
Más detalles con n→0 a la derecha
$$\lim_{n \to 1^-}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = \frac{-1 + e^{2 \pi}}{- e^{\frac{\pi}{2}} + e^{\frac{3 \pi}{2}}}$$
Más detalles con n→1 a la izquierda
$$\lim_{n \to 1^+}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = \frac{-1 + e^{2 \pi}}{- e^{\frac{\pi}{2}} + e^{\frac{3 \pi}{2}}}$$
Más detalles con n→1 a la derecha
$$\lim_{n \to -\infty}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = 1$$
Más detalles con n→-oo
$$\lim_{n \to 0^-}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = -\infty$$
Más detalles con n→0 a la izquierda
$$\lim_{n \to 0^+}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = \infty$$
Más detalles con n→0 a la derecha
$$\lim_{n \to 1^-}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = \frac{-1 + e^{2 \pi}}{- e^{\frac{\pi}{2}} + e^{\frac{3 \pi}{2}}}$$
Más detalles con n→1 a la izquierda
$$\lim_{n \to 1^+}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = \frac{-1 + e^{2 \pi}}{- e^{\frac{\pi}{2}} + e^{\frac{3 \pi}{2}}}$$
Más detalles con n→1 a la derecha
$$\lim_{n \to -\infty}\left(\frac{\sinh{\left(\frac{\pi}{n} \right)}}{\sinh{\left(\frac{\pi}{n + 1} \right)}}\right) = 1$$
Más detalles con n→-oo