Sr Examen
Otras calculadoras:
-
¿Cómo usar?
- Límite de la función:
-
Límite de (9+x^2-6*x)/(x^2-3*x)
-
Límite de 1+1/x
-
Límite de (-2+sqrt(-2+x))/(-6+x)
-
Límite de (sqrt(12+x)-sqrt(4-x))/(-8+x^2+2*x)
-
Expresiones idénticas
- -x/(x-sin(x))+tan(x)
- menos x dividir por (x menos seno de (x)) más tangente de (x)
- -x/x-sinx+tanx
- -x dividir por (x-sin(x))+tan(x)
-
Expresiones semejantes
- -x/(x+sin(x))+tan(x)
- -x/(x-sin(x))-tan(x)
- x/(x-sin(x))+tan(x)
- -x/(x-sinx)+tan(x)
-
Expresiones con funciones
- Seno sin
- sin(x)^2/(1+cos(x))
- sin(3*x)/(9*x)
- sin(3*x)/x^2
- sin(3*x)/sin(x)
- sin(x)/(x+tan(x))
- Tangente tan
- tan(2*x)/(x^2-x)
- tan(x+pi/3)^(2+x)
- tan(8*x)/(6*x)
- tan(2*x)/sin(x)
- tan(x)^2/sin(2*x)
Límite de la función -x/(x-sin(x))+tan(x)
Solución
Ha introducido
[src]
/ -x \ lim |---------- + tan(x)| x->0+\x - sin(x) /
$$\lim_{x \to 0^+}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right)$$
Limit((-x)/(x - sin(x)) + tan(x), x, 0)
Método de l'Hopital
Tenemos la indeterminación de tipo
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(x \tan{\left(x \right)} - x - \sin{\left(x \right)} \tan{\left(x \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+}\left(x - \sin{\left(x \right)}\right) = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 0^+}\left(\frac{- x + \left(x - \sin{\left(x \right)}\right) \tan{\left(x \right)}}{x - \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x \tan{\left(x \right)} - x - \sin{\left(x \right)} \tan{\left(x \right)}\right)}{\frac{d}{d x} \left(x - \sin{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x \tan^{2}{\left(x \right)} + x - \sin{\left(x \right)} \tan^{2}{\left(x \right)} - \sin{\left(x \right)} - \cos{\left(x \right)} \tan{\left(x \right)} + \tan{\left(x \right)} - 1}{1 - \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x \tan^{2}{\left(x \right)} + x - \sin{\left(x \right)} \tan^{2}{\left(x \right)} - \sin{\left(x \right)} - \cos{\left(x \right)} \tan{\left(x \right)} + \tan{\left(x \right)} - 1\right)}{\frac{d}{d x} \left(1 - \cos{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 x \tan^{3}{\left(x \right)} + 2 x \tan{\left(x \right)} - 2 \sin{\left(x \right)} \tan^{3}{\left(x \right)} - \sin{\left(x \right)} \tan{\left(x \right)} - 2 \cos{\left(x \right)} \tan^{2}{\left(x \right)} - 2 \cos{\left(x \right)} + 2 \tan^{2}{\left(x \right)} + 2}{\sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 x \tan^{3}{\left(x \right)} + 2 x \tan{\left(x \right)} - 2 \sin{\left(x \right)} \tan^{3}{\left(x \right)} - \sin{\left(x \right)} \tan{\left(x \right)} - 2 \cos{\left(x \right)} \tan^{2}{\left(x \right)} - 2 \cos{\left(x \right)} + 2 \tan^{2}{\left(x \right)} + 2\right)}{\frac{d}{d x} \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{6 x \tan^{4}{\left(x \right)} + 8 x \tan^{2}{\left(x \right)} + 2 x - 6 \sin{\left(x \right)} \tan^{4}{\left(x \right)} - 5 \sin{\left(x \right)} \tan^{2}{\left(x \right)} + \sin{\left(x \right)} - 6 \cos{\left(x \right)} \tan^{3}{\left(x \right)} - 5 \cos{\left(x \right)} \tan{\left(x \right)} + 6 \tan^{3}{\left(x \right)} + 6 \tan{\left(x \right)}}{\cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{6 x \tan^{4}{\left(x \right)} + 8 x \tan^{2}{\left(x \right)} + 2 x - 6 \sin{\left(x \right)} \tan^{4}{\left(x \right)} - 5 \sin{\left(x \right)} \tan^{2}{\left(x \right)} + \sin{\left(x \right)} - 6 \cos{\left(x \right)} \tan^{3}{\left(x \right)} - 5 \cos{\left(x \right)} \tan{\left(x \right)} + 6 \tan^{3}{\left(x \right)} + 6 \tan{\left(x \right)}}{\cos{\left(x \right)}}\right)$$
=
$$-\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(x \tan{\left(x \right)} - x - \sin{\left(x \right)} \tan{\left(x \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+}\left(x - \sin{\left(x \right)}\right) = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 0^+}\left(\frac{- x + \left(x - \sin{\left(x \right)}\right) \tan{\left(x \right)}}{x - \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x \tan{\left(x \right)} - x - \sin{\left(x \right)} \tan{\left(x \right)}\right)}{\frac{d}{d x} \left(x - \sin{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x \tan^{2}{\left(x \right)} + x - \sin{\left(x \right)} \tan^{2}{\left(x \right)} - \sin{\left(x \right)} - \cos{\left(x \right)} \tan{\left(x \right)} + \tan{\left(x \right)} - 1}{1 - \cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x \tan^{2}{\left(x \right)} + x - \sin{\left(x \right)} \tan^{2}{\left(x \right)} - \sin{\left(x \right)} - \cos{\left(x \right)} \tan{\left(x \right)} + \tan{\left(x \right)} - 1\right)}{\frac{d}{d x} \left(1 - \cos{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 x \tan^{3}{\left(x \right)} + 2 x \tan{\left(x \right)} - 2 \sin{\left(x \right)} \tan^{3}{\left(x \right)} - \sin{\left(x \right)} \tan{\left(x \right)} - 2 \cos{\left(x \right)} \tan^{2}{\left(x \right)} - 2 \cos{\left(x \right)} + 2 \tan^{2}{\left(x \right)} + 2}{\sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 x \tan^{3}{\left(x \right)} + 2 x \tan{\left(x \right)} - 2 \sin{\left(x \right)} \tan^{3}{\left(x \right)} - \sin{\left(x \right)} \tan{\left(x \right)} - 2 \cos{\left(x \right)} \tan^{2}{\left(x \right)} - 2 \cos{\left(x \right)} + 2 \tan^{2}{\left(x \right)} + 2\right)}{\frac{d}{d x} \sin{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{6 x \tan^{4}{\left(x \right)} + 8 x \tan^{2}{\left(x \right)} + 2 x - 6 \sin{\left(x \right)} \tan^{4}{\left(x \right)} - 5 \sin{\left(x \right)} \tan^{2}{\left(x \right)} + \sin{\left(x \right)} - 6 \cos{\left(x \right)} \tan^{3}{\left(x \right)} - 5 \cos{\left(x \right)} \tan{\left(x \right)} + 6 \tan^{3}{\left(x \right)} + 6 \tan{\left(x \right)}}{\cos{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{6 x \tan^{4}{\left(x \right)} + 8 x \tan^{2}{\left(x \right)} + 2 x - 6 \sin{\left(x \right)} \tan^{4}{\left(x \right)} - 5 \sin{\left(x \right)} \tan^{2}{\left(x \right)} + \sin{\left(x \right)} - 6 \cos{\left(x \right)} \tan^{3}{\left(x \right)} - 5 \cos{\left(x \right)} \tan{\left(x \right)} + 6 \tan^{3}{\left(x \right)} + 6 \tan{\left(x \right)}}{\cos{\left(x \right)}}\right)$$
=
$$-\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)
Gráfica
A la izquierda y a la derecha
[src]
/ -x \ lim |---------- + tan(x)| x->0+\x - sin(x) /
$$\lim_{x \to 0^+}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right)$$
-oo
$$-\infty$$
= -136806.293377731
/ -x \ lim |---------- + tan(x)| x->0-\x - sin(x) /
$$\lim_{x \to 0^-}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right)$$
-oo
$$-\infty$$
= -136806.306622958
= -136806.306622958
Otros límites con x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right) = -\infty$$
Más detalles con x→0 a la izquierda
$$\lim_{x \to 0^+}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right) = -\infty$$
$$\lim_{x \to \infty}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right)$$
Más detalles con x→oo
$$\lim_{x \to 1^-}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right) = \frac{- \tan{\left(1 \right)} + 1 + \sin{\left(1 \right)} \tan{\left(1 \right)}}{-1 + \sin{\left(1 \right)}}$$
Más detalles con x→1 a la izquierda
$$\lim_{x \to 1^+}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right) = \frac{- \tan{\left(1 \right)} + 1 + \sin{\left(1 \right)} \tan{\left(1 \right)}}{-1 + \sin{\left(1 \right)}}$$
Más detalles con x→1 a la derecha
$$\lim_{x \to -\infty}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right)$$
Más detalles con x→-oo
Más detalles con x→0 a la izquierda
$$\lim_{x \to 0^+}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right) = -\infty$$
$$\lim_{x \to \infty}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right)$$
Más detalles con x→oo
$$\lim_{x \to 1^-}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right) = \frac{- \tan{\left(1 \right)} + 1 + \sin{\left(1 \right)} \tan{\left(1 \right)}}{-1 + \sin{\left(1 \right)}}$$
Más detalles con x→1 a la izquierda
$$\lim_{x \to 1^+}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right) = \frac{- \tan{\left(1 \right)} + 1 + \sin{\left(1 \right)} \tan{\left(1 \right)}}{-1 + \sin{\left(1 \right)}}$$
Más detalles con x→1 a la derecha
$$\lim_{x \to -\infty}\left(\frac{\left(-1\right) x}{x - \sin{\left(x \right)}} + \tan{\left(x \right)}\right)$$
Más detalles con x→-oo
Gráfico