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Expresiones idénticas
- (-x/((uno -x/log(nueve))*log(nueve)))^x
- ( menos x dividir por ((1 menos x dividir por logaritmo de (9)) multiplicar por logaritmo de (9))) en el grado x
- ( menos x dividir por ((uno menos x dividir por logaritmo de (nueve)) multiplicar por logaritmo de (nueve))) en el grado x
- (-x/((1-x/log(9))*log(9)))x
- -x/1-x/log9*log9x
- (-x/((1-x/log(9))log(9)))^x
- (-x/((1-x/log(9))log(9)))x
- -x/1-x/log9log9x
- -x/1-x/log9log9^x
- (-x dividir por ((1-x dividir por log(9))*log(9)))^x
-
Expresiones semejantes
- (x/((1-x/log(9))*log(9)))^x
- (-x/((1+x/log(9))*log(9)))^x
-
Expresiones con funciones
- Logaritmo log
- log(1+k*x)/x
- log(sin(3*x))/log(sin(5*x))
- log(1+3*x)
- log(2+n)^2*(2+n)/((1+n)*log(1+n)^2)
- log(2+sqrt(atan(x)*sin(1/x)))
- Logaritmo log
- log(1+k*x)/x
- log(sin(3*x))/log(sin(5*x))
- log(1+3*x)
- log(2+n)^2*(2+n)/((1+n)*log(1+n)^2)
- log(2+sqrt(atan(x)*sin(1/x)))
Límite de la función (-x/((1-x/log(9))*log(9)))^x
Solución
Ha introducido
[src]
x
/ -x \
lim |-------------------|
x->oo|/ x \ |
||1 - ------|*log(9)|
\\ log(9)/ /
$$\lim_{x \to \infty} \left(\frac{\left(-1\right) x}{\left(- \frac{x}{\log{\left(9 \right)}} + 1\right) \log{\left(9 \right)}}\right)^{x}$$
Limit(((-x)/(((1 - x/log(9))*log(9))))^x, x, oo, dir='-')
Solución detallada
Tomamos como el límite
$$\lim_{x \to \infty} \left(\frac{\left(-1\right) x}{\left(- \frac{x}{\log{\left(9 \right)}} + 1\right) \log{\left(9 \right)}}\right)^{x}$$
cambiamos
$$\lim_{x \to \infty} \left(\frac{\left(-1\right) x}{\left(- \frac{x}{\log{\left(9 \right)}} + 1\right) \log{\left(9 \right)}}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(- \frac{x}{\log{\left(9 \right)}} + 1\right) - 1}{- \frac{x}{\log{\left(9 \right)}} + 1}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(- \frac{1}{- \frac{x}{\log{\left(9 \right)}} + 1} + \frac{- \frac{x}{\log{\left(9 \right)}} + 1}{- \frac{x}{\log{\left(9 \right)}} + 1}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{1}{- \frac{x}{\log{\left(9 \right)}} + 1}\right)^{x}$$
=
hacemos el cambio
$$u = \frac{- \frac{x}{\log{\left(9 \right)}} + 1}{-1}$$
entonces
$$\lim_{x \to \infty} \left(1 - \frac{1}{- \frac{x}{\log{\left(9 \right)}} + 1}\right)^{x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\left(u + 1\right) \log{\left(9 \right)}}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{u \frac{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}{u}} \left(1 + \frac{1}{u}\right)^{\log{\left(9 \right)}}\right)$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\log{\left(9 \right)}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}{u}}$$
El límite
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
hay el segundo límite, es igual a e ~ 2.718281828459045
entonces
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}{u}} = e^{\frac{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}{u}}$$
Entonces la respuesta definitiva es:
$$\lim_{x \to \infty} \left(\frac{\left(-1\right) x}{\left(- \frac{x}{\log{\left(9 \right)}} + 1\right) \log{\left(9 \right)}}\right)^{x} = 9$$
$$\lim_{x \to \infty} \left(\frac{\left(-1\right) x}{\left(- \frac{x}{\log{\left(9 \right)}} + 1\right) \log{\left(9 \right)}}\right)^{x}$$
cambiamos
$$\lim_{x \to \infty} \left(\frac{\left(-1\right) x}{\left(- \frac{x}{\log{\left(9 \right)}} + 1\right) \log{\left(9 \right)}}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(\frac{\left(- \frac{x}{\log{\left(9 \right)}} + 1\right) - 1}{- \frac{x}{\log{\left(9 \right)}} + 1}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(- \frac{1}{- \frac{x}{\log{\left(9 \right)}} + 1} + \frac{- \frac{x}{\log{\left(9 \right)}} + 1}{- \frac{x}{\log{\left(9 \right)}} + 1}\right)^{x}$$
=
$$\lim_{x \to \infty} \left(1 - \frac{1}{- \frac{x}{\log{\left(9 \right)}} + 1}\right)^{x}$$
=
hacemos el cambio
$$u = \frac{- \frac{x}{\log{\left(9 \right)}} + 1}{-1}$$
entonces
$$\lim_{x \to \infty} \left(1 - \frac{1}{- \frac{x}{\log{\left(9 \right)}} + 1}\right)^{x}$$ =
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\left(u + 1\right) \log{\left(9 \right)}}$$
=
$$\lim_{u \to \infty}\left(\left(1 + \frac{1}{u}\right)^{u \frac{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}{u}} \left(1 + \frac{1}{u}\right)^{\log{\left(9 \right)}}\right)$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\log{\left(9 \right)}} \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}$$
=
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}$$
=
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}{u}}$$
El límite
$$\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}$$
hay el segundo límite, es igual a e ~ 2.718281828459045
entonces
$$\left(\left(\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{u}\right)\right)^{\frac{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}{u}} = e^{\frac{\left(u + 1\right) \log{\left(9 \right)} - \log{\left(9 \right)}}{u}}$$
Entonces la respuesta definitiva es:
$$\lim_{x \to \infty} \left(\frac{\left(-1\right) x}{\left(- \frac{x}{\log{\left(9 \right)}} + 1\right) \log{\left(9 \right)}}\right)^{x} = 9$$
Método de l'Hopital
En el caso de esta función, no tiene sentido aplicar el Método de l'Hopital, ya que no existe la indeterminación tipo 0/0 or oo/oo
Gráfica