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Otras calculadoras:
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- Límite de la función:
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Límite de (7^(2*x)-5^(3*x))/(-atan(3*x)+2*x)
-
Límite de (-2+sqrt(3+x))/(-1+x)
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Límite de (-15+x+2*x^2)/(-6+3*x^2+7*x)
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Límite de 1/(-5+x)
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Expresiones idénticas
- log(cuatro + tres *n)^ dos *(uno +n)/(n*log(uno + tres *n)^ dos)
- logaritmo de (4 más 3 multiplicar por n) al cuadrado multiplicar por (1 más n) dividir por (n multiplicar por logaritmo de (1 más 3 multiplicar por n) al cuadrado )
- logaritmo de (cuatro más tres multiplicar por n) en el grado dos multiplicar por (uno más n) dividir por (n multiplicar por logaritmo de (uno más tres multiplicar por n) en el grado dos)
- log(4+3*n)2*(1+n)/(n*log(1+3*n)2)
- log4+3*n2*1+n/n*log1+3*n2
- log(4+3*n)²*(1+n)/(n*log(1+3*n)²)
- log(4+3*n) en el grado 2*(1+n)/(n*log(1+3*n) en el grado 2)
- log(4+3n)^2(1+n)/(nlog(1+3n)^2)
- log(4+3n)2(1+n)/(nlog(1+3n)2)
- log4+3n21+n/nlog1+3n2
- log4+3n^21+n/nlog1+3n^2
- log(4+3*n)^2*(1+n) dividir por (n*log(1+3*n)^2)
-
Expresiones semejantes
- log(4+3*n)^2*(1+n)/(n*log(1-3*n)^2)
- log(4+3*n)^2*(1-n)/(n*log(1+3*n)^2)
- log(4-3*n)^2*(1+n)/(n*log(1+3*n)^2)
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Expresiones con funciones
- Logaritmo log
- log(sqrt((1+x)/(1-x)))/x
- log(x-a)/log(e^x-e^a)
- log(5+x)/(3+x)^(1/4)
- log(sin(3*x))/(-pi+6*x)^2
- log(1-3*x)/(-1+e^(4*x))
- Logaritmo log
- log(sqrt((1+x)/(1-x)))/x
- log(x-a)/log(e^x-e^a)
- log(5+x)/(3+x)^(1/4)
- log(sin(3*x))/(-pi+6*x)^2
- log(1-3*x)/(-1+e^(4*x))
Límite de la función log(4+3*n)^2*(1+n)/(n*log(1+3*n)^2)
Solución
Ha introducido
[src]
/ 2 \
|log (4 + 3*n)*(1 + n)|
lim |---------------------|
n->oo| 2 |
\ n*log (1 + 3*n) /
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right)$$
Limit((log(4 + 3*n)^2*(1 + n))/((n*log(1 + 3*n)^2)), n, oo, dir='-')
Método de l'Hopital
Tenemos la indeterminación de tipo
tal que el límite para el numerador es
$$\lim_{n \to \infty} \log{\left(3 n + 4 \right)}^{2} = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty}\left(\frac{n \log{\left(3 n + 1 \right)}^{2}}{n + 1}\right) = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \log{\left(3 n + 4 \right)}^{2}}{\frac{d}{d n} \frac{n \log{\left(3 n + 1 \right)}^{2}}{n + 1}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{1}{\left(\frac{n}{2 \log{\left(3 n + 4 \right)}} + \frac{2}{3 \log{\left(3 n + 4 \right)}}\right) \left(\frac{6 n \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} - \frac{n \log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n + 1}\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{1}{\frac{6 n \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} - \frac{n \log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n + 1}}}{\frac{d}{d n} \left(\frac{n}{2 \log{\left(3 n + 4 \right)}} + \frac{2}{3 \log{\left(3 n + 4 \right)}}\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{- \frac{6 n \left(- 6 n - 4\right) \log{\left(3 n + 1 \right)}}{\left(3 n^{2} + 4 n + 1\right)^{2}} + \frac{n \left(- 2 n - 2\right) \log{\left(3 n + 1 \right)}^{2}}{\left(n^{2} + 2 n + 1\right)^{2}} - \frac{18 n}{\left(3 n + 1\right) \left(3 n^{2} + 4 n + 1\right)} + \frac{6 n \log{\left(3 n + 1 \right)}}{\left(3 n + 1\right) \left(n^{2} + 2 n + 1\right)} - \frac{6 \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} - \frac{6 \log{\left(3 n + 1 \right)}}{\left(n + 1\right) \left(3 n + 1\right)} + \frac{\log{\left(3 n + 1 \right)}^{2}}{\left(n + 1\right)^{2}}}{\left(- \frac{3 n}{2 \left(3 n + 4\right) \log{\left(3 n + 4 \right)}^{2}} + \frac{1}{2 \log{\left(3 n + 4 \right)}} - \frac{2}{\left(3 n + 4\right) \log{\left(3 n + 4 \right)}^{2}}\right) \left(\frac{6 n \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} - \frac{n \log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n + 1}\right)^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{- \frac{6 n \left(- 6 n - 4\right) \log{\left(3 n + 1 \right)}}{\left(3 n^{2} + 4 n + 1\right)^{2}} + \frac{n \left(- 2 n - 2\right) \log{\left(3 n + 1 \right)}^{2}}{\left(n^{2} + 2 n + 1\right)^{2}} - \frac{18 n}{\left(3 n + 1\right) \left(3 n^{2} + 4 n + 1\right)} + \frac{6 n \log{\left(3 n + 1 \right)}}{\left(3 n + 1\right) \left(n^{2} + 2 n + 1\right)} - \frac{6 \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} - \frac{6 \log{\left(3 n + 1 \right)}}{\left(n + 1\right) \left(3 n + 1\right)} + \frac{\log{\left(3 n + 1 \right)}^{2}}{\left(n + 1\right)^{2}}}{\left(- \frac{3 n}{2 \left(3 n + 4\right) \log{\left(3 n + 4 \right)}^{2}} + \frac{1}{2 \log{\left(3 n + 4 \right)}} - \frac{2}{\left(3 n + 4\right) \log{\left(3 n + 4 \right)}^{2}}\right) \left(\frac{6 n \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} - \frac{n \log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n + 1}\right)^{2}}\right)$$
=
$$1$$
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
$$\lim_{n \to \infty} \log{\left(3 n + 4 \right)}^{2} = \infty$$
y el límite para el denominador es
$$\lim_{n \to \infty}\left(\frac{n \log{\left(3 n + 1 \right)}^{2}}{n + 1}\right) = \infty$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \log{\left(3 n + 4 \right)}^{2}}{\frac{d}{d n} \frac{n \log{\left(3 n + 1 \right)}^{2}}{n + 1}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{1}{\left(\frac{n}{2 \log{\left(3 n + 4 \right)}} + \frac{2}{3 \log{\left(3 n + 4 \right)}}\right) \left(\frac{6 n \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} - \frac{n \log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n + 1}\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \frac{1}{\frac{6 n \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} - \frac{n \log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n + 1}}}{\frac{d}{d n} \left(\frac{n}{2 \log{\left(3 n + 4 \right)}} + \frac{2}{3 \log{\left(3 n + 4 \right)}}\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{- \frac{6 n \left(- 6 n - 4\right) \log{\left(3 n + 1 \right)}}{\left(3 n^{2} + 4 n + 1\right)^{2}} + \frac{n \left(- 2 n - 2\right) \log{\left(3 n + 1 \right)}^{2}}{\left(n^{2} + 2 n + 1\right)^{2}} - \frac{18 n}{\left(3 n + 1\right) \left(3 n^{2} + 4 n + 1\right)} + \frac{6 n \log{\left(3 n + 1 \right)}}{\left(3 n + 1\right) \left(n^{2} + 2 n + 1\right)} - \frac{6 \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} - \frac{6 \log{\left(3 n + 1 \right)}}{\left(n + 1\right) \left(3 n + 1\right)} + \frac{\log{\left(3 n + 1 \right)}^{2}}{\left(n + 1\right)^{2}}}{\left(- \frac{3 n}{2 \left(3 n + 4\right) \log{\left(3 n + 4 \right)}^{2}} + \frac{1}{2 \log{\left(3 n + 4 \right)}} - \frac{2}{\left(3 n + 4\right) \log{\left(3 n + 4 \right)}^{2}}\right) \left(\frac{6 n \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} - \frac{n \log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n + 1}\right)^{2}}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{- \frac{6 n \left(- 6 n - 4\right) \log{\left(3 n + 1 \right)}}{\left(3 n^{2} + 4 n + 1\right)^{2}} + \frac{n \left(- 2 n - 2\right) \log{\left(3 n + 1 \right)}^{2}}{\left(n^{2} + 2 n + 1\right)^{2}} - \frac{18 n}{\left(3 n + 1\right) \left(3 n^{2} + 4 n + 1\right)} + \frac{6 n \log{\left(3 n + 1 \right)}}{\left(3 n + 1\right) \left(n^{2} + 2 n + 1\right)} - \frac{6 \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} - \frac{6 \log{\left(3 n + 1 \right)}}{\left(n + 1\right) \left(3 n + 1\right)} + \frac{\log{\left(3 n + 1 \right)}^{2}}{\left(n + 1\right)^{2}}}{\left(- \frac{3 n}{2 \left(3 n + 4\right) \log{\left(3 n + 4 \right)}^{2}} + \frac{1}{2 \log{\left(3 n + 4 \right)}} - \frac{2}{\left(3 n + 4\right) \log{\left(3 n + 4 \right)}^{2}}\right) \left(\frac{6 n \log{\left(3 n + 1 \right)}}{3 n^{2} + 4 n + 1} - \frac{n \log{\left(3 n + 1 \right)}^{2}}{n^{2} + 2 n + 1} + \frac{\log{\left(3 n + 1 \right)}^{2}}{n + 1}\right)^{2}}\right)$$
=
$$1$$
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
$$\lim_{n \to \infty}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = -\infty$$
Más detalles con n→0 a la izquierda
$$\lim_{n \to 0^+}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = \infty$$
Más detalles con n→0 a la derecha
$$\lim_{n \to 1^-}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = \frac{\log{\left(7 \right)}^{2}}{2 \log{\left(2 \right)}^{2}}$$
Más detalles con n→1 a la izquierda
$$\lim_{n \to 1^+}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = \frac{\log{\left(7 \right)}^{2}}{2 \log{\left(2 \right)}^{2}}$$
Más detalles con n→1 a la derecha
$$\lim_{n \to -\infty}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = 1$$
Más detalles con n→-oo
$$\lim_{n \to 0^-}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = -\infty$$
Más detalles con n→0 a la izquierda
$$\lim_{n \to 0^+}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = \infty$$
Más detalles con n→0 a la derecha
$$\lim_{n \to 1^-}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = \frac{\log{\left(7 \right)}^{2}}{2 \log{\left(2 \right)}^{2}}$$
Más detalles con n→1 a la izquierda
$$\lim_{n \to 1^+}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = \frac{\log{\left(7 \right)}^{2}}{2 \log{\left(2 \right)}^{2}}$$
Más detalles con n→1 a la derecha
$$\lim_{n \to -\infty}\left(\frac{\left(n + 1\right) \log{\left(3 n + 4 \right)}^{2}}{n \log{\left(3 n + 1 \right)}^{2}}\right) = 1$$
Más detalles con n→-oo
Gráfico