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- Límite de la función:
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Límite de (8+x)/x^2
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Límite de (-1+4*x+5*x^2)/(-2+x+3*x^2)
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Límite de (-5-14*x+3*x^2)/(-15+x^2-2*x)
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Límite de (-40-2*x+3*x^2)/(-4+x^2-3*x)
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Expresiones idénticas
- sin(x)^ dos -tan(x)^ dos /x^ cuatro
- seno de (x) al cuadrado menos tangente de (x) al cuadrado dividir por x en el grado 4
- seno de (x) en el grado dos menos tangente de (x) en el grado dos dividir por x en el grado cuatro
- sin(x)2-tan(x)2/x4
- sinx2-tanx2/x4
- sin(x)²-tan(x)²/x⁴
- sin(x) en el grado 2-tan(x) en el grado 2/x en el grado 4
- sinx^2-tanx^2/x^4
- sin(x)^2-tan(x)^2 dividir por x^4
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Expresiones semejantes
- sin(x)^2+tan(x)^2/x^4
- sinx^2-tan(x)^2/x^4
-
Expresiones con funciones
- Seno sin
- sin(x)^(2/(3+log(x)))
- sin(x)^2/(2*x^2)
- sin(8*x)/tan(4*x)
- sin(8*x)/cos(7*x)
- sin(8*x)^2/(2*x^2)
- Tangente tan
- tan(1+x)/(-1+x^2)
- tan(-1+x)/(-1+x)
- tan(6*x)/sin(2*x)
- tan(4*x)/(7*x^2)
- tan(3)^2/(10*x)
Límite de la función sin(x)^2-tan(x)^2/x^4
Solución
Ha introducido
[src]
/ 2 \
| 2 tan (x)|
lim |sin (x) - -------|
x->0+| 4 |
\ x /
$$\lim_{x \to 0^+}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right)$$
Limit(sin(x)^2 - tan(x)^2/x^4, x, 0)
Método de l'Hopital
Tenemos la indeterminación de tipo
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(x^{4} \sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} x^{4} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 0^+}\left(\frac{x^{4} \sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}}{x^{4}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x^{4} \sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}\right)}{\frac{d}{d x} x^{4}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 x^{4} \sin{\left(x \right)} \cos{\left(x \right)} + 4 x^{3} \sin^{2}{\left(x \right)} - 2 \tan^{3}{\left(x \right)} - 2 \tan{\left(x \right)}}{4 x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 x^{4} \sin{\left(x \right)} \cos{\left(x \right)} + 4 x^{3} \sin^{2}{\left(x \right)} - 2 \tan^{3}{\left(x \right)} - 2 \tan{\left(x \right)}\right)}{\frac{d}{d x} 4 x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 2 x^{4} \sin^{2}{\left(x \right)} + 2 x^{4} \cos^{2}{\left(x \right)} + 16 x^{3} \sin{\left(x \right)} \cos{\left(x \right)} + 12 x^{2} \sin^{2}{\left(x \right)} - 6 \tan^{4}{\left(x \right)} - 8 \tan^{2}{\left(x \right)} - 2}{12 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 2 x^{4} \sin^{2}{\left(x \right)} + 2 x^{4} \cos^{2}{\left(x \right)} + 16 x^{3} \sin{\left(x \right)} \cos{\left(x \right)} + 12 x^{2} \sin^{2}{\left(x \right)} - 6 \tan^{4}{\left(x \right)} - 8 \tan^{2}{\left(x \right)} - 2\right)}{\frac{d}{d x} 12 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 8 x^{4} \sin{\left(x \right)} \cos{\left(x \right)} - 24 x^{3} \sin^{2}{\left(x \right)} + 24 x^{3} \cos^{2}{\left(x \right)} + 72 x^{2} \sin{\left(x \right)} \cos{\left(x \right)} + 24 x \sin^{2}{\left(x \right)} - 24 \tan^{5}{\left(x \right)} - 40 \tan^{3}{\left(x \right)} - 16 \tan{\left(x \right)}}{24 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 8 x^{4} \sin{\left(x \right)} \cos{\left(x \right)} - 24 x^{3} \sin^{2}{\left(x \right)} + 24 x^{3} \cos^{2}{\left(x \right)} + 72 x^{2} \sin{\left(x \right)} \cos{\left(x \right)} + 24 x \sin^{2}{\left(x \right)} - 24 \tan^{5}{\left(x \right)} - 40 \tan^{3}{\left(x \right)} - 16 \tan{\left(x \right)}\right)}{\frac{d}{d x} 24 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x^{4} \sin^{2}{\left(x \right)}}{3} - \frac{x^{4} \cos^{2}{\left(x \right)}}{3} - \frac{16 x^{3} \sin{\left(x \right)} \cos{\left(x \right)}}{3} - 6 x^{2} \sin^{2}{\left(x \right)} + 6 x^{2} \cos^{2}{\left(x \right)} + 8 x \sin{\left(x \right)} \cos{\left(x \right)} + \sin^{2}{\left(x \right)} - 5 \tan^{6}{\left(x \right)} - 10 \tan^{4}{\left(x \right)} - \frac{17 \tan^{2}{\left(x \right)}}{3} - \frac{2}{3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x^{4} \sin^{2}{\left(x \right)}}{3} - \frac{x^{4} \cos^{2}{\left(x \right)}}{3} - \frac{16 x^{3} \sin{\left(x \right)} \cos{\left(x \right)}}{3} - 6 x^{2} \sin^{2}{\left(x \right)} + 6 x^{2} \cos^{2}{\left(x \right)} + 8 x \sin{\left(x \right)} \cos{\left(x \right)} + \sin^{2}{\left(x \right)} - 5 \tan^{6}{\left(x \right)} - 10 \tan^{4}{\left(x \right)} - \frac{17 \tan^{2}{\left(x \right)}}{3} - \frac{2}{3}\right)$$
=
$$-\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 4 vez (veces)
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(x^{4} \sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} x^{4} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 0^+}\left(\frac{x^{4} \sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}}{x^{4}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(x^{4} \sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}\right)}{\frac{d}{d x} x^{4}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{2 x^{4} \sin{\left(x \right)} \cos{\left(x \right)} + 4 x^{3} \sin^{2}{\left(x \right)} - 2 \tan^{3}{\left(x \right)} - 2 \tan{\left(x \right)}}{4 x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 x^{4} \sin{\left(x \right)} \cos{\left(x \right)} + 4 x^{3} \sin^{2}{\left(x \right)} - 2 \tan^{3}{\left(x \right)} - 2 \tan{\left(x \right)}\right)}{\frac{d}{d x} 4 x^{3}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 2 x^{4} \sin^{2}{\left(x \right)} + 2 x^{4} \cos^{2}{\left(x \right)} + 16 x^{3} \sin{\left(x \right)} \cos{\left(x \right)} + 12 x^{2} \sin^{2}{\left(x \right)} - 6 \tan^{4}{\left(x \right)} - 8 \tan^{2}{\left(x \right)} - 2}{12 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 2 x^{4} \sin^{2}{\left(x \right)} + 2 x^{4} \cos^{2}{\left(x \right)} + 16 x^{3} \sin{\left(x \right)} \cos{\left(x \right)} + 12 x^{2} \sin^{2}{\left(x \right)} - 6 \tan^{4}{\left(x \right)} - 8 \tan^{2}{\left(x \right)} - 2\right)}{\frac{d}{d x} 12 x^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- 8 x^{4} \sin{\left(x \right)} \cos{\left(x \right)} - 24 x^{3} \sin^{2}{\left(x \right)} + 24 x^{3} \cos^{2}{\left(x \right)} + 72 x^{2} \sin{\left(x \right)} \cos{\left(x \right)} + 24 x \sin^{2}{\left(x \right)} - 24 \tan^{5}{\left(x \right)} - 40 \tan^{3}{\left(x \right)} - 16 \tan{\left(x \right)}}{24 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 8 x^{4} \sin{\left(x \right)} \cos{\left(x \right)} - 24 x^{3} \sin^{2}{\left(x \right)} + 24 x^{3} \cos^{2}{\left(x \right)} + 72 x^{2} \sin{\left(x \right)} \cos{\left(x \right)} + 24 x \sin^{2}{\left(x \right)} - 24 \tan^{5}{\left(x \right)} - 40 \tan^{3}{\left(x \right)} - 16 \tan{\left(x \right)}\right)}{\frac{d}{d x} 24 x}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x^{4} \sin^{2}{\left(x \right)}}{3} - \frac{x^{4} \cos^{2}{\left(x \right)}}{3} - \frac{16 x^{3} \sin{\left(x \right)} \cos{\left(x \right)}}{3} - 6 x^{2} \sin^{2}{\left(x \right)} + 6 x^{2} \cos^{2}{\left(x \right)} + 8 x \sin{\left(x \right)} \cos{\left(x \right)} + \sin^{2}{\left(x \right)} - 5 \tan^{6}{\left(x \right)} - 10 \tan^{4}{\left(x \right)} - \frac{17 \tan^{2}{\left(x \right)}}{3} - \frac{2}{3}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{x^{4} \sin^{2}{\left(x \right)}}{3} - \frac{x^{4} \cos^{2}{\left(x \right)}}{3} - \frac{16 x^{3} \sin{\left(x \right)} \cos{\left(x \right)}}{3} - 6 x^{2} \sin^{2}{\left(x \right)} + 6 x^{2} \cos^{2}{\left(x \right)} + 8 x \sin{\left(x \right)} \cos{\left(x \right)} + \sin^{2}{\left(x \right)} - 5 \tan^{6}{\left(x \right)} - 10 \tan^{4}{\left(x \right)} - \frac{17 \tan^{2}{\left(x \right)}}{3} - \frac{2}{3}\right)$$
=
$$-\infty$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 4 vez (veces)
Gráfica
A la izquierda y a la derecha
[src]
/ 2 \
| 2 tan (x)|
lim |sin (x) - -------|
x->0+| 4 |
\ x /
$$\lim_{x \to 0^+}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right)$$
-oo
$$-\infty$$
= -22801.6666393784
/ 2 \
| 2 tan (x)|
lim |sin (x) - -------|
x->0-| 4 |
\ x /
$$\lim_{x \to 0^-}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right)$$
-oo
$$-\infty$$
= -22801.6666393784
= -22801.6666393784
Otros límites con x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right) = -\infty$$
Más detalles con x→0 a la izquierda
$$\lim_{x \to 0^+}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right) = -\infty$$
$$\lim_{x \to \infty}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right)$$
Más detalles con x→oo
$$\lim_{x \to 1^-}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right) = - \tan^{2}{\left(1 \right)} + \sin^{2}{\left(1 \right)}$$
Más detalles con x→1 a la izquierda
$$\lim_{x \to 1^+}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right) = - \tan^{2}{\left(1 \right)} + \sin^{2}{\left(1 \right)}$$
Más detalles con x→1 a la derecha
$$\lim_{x \to -\infty}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right)$$
Más detalles con x→-oo
Más detalles con x→0 a la izquierda
$$\lim_{x \to 0^+}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right) = -\infty$$
$$\lim_{x \to \infty}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right)$$
Más detalles con x→oo
$$\lim_{x \to 1^-}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right) = - \tan^{2}{\left(1 \right)} + \sin^{2}{\left(1 \right)}$$
Más detalles con x→1 a la izquierda
$$\lim_{x \to 1^+}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right) = - \tan^{2}{\left(1 \right)} + \sin^{2}{\left(1 \right)}$$
Más detalles con x→1 a la derecha
$$\lim_{x \to -\infty}\left(\sin^{2}{\left(x \right)} - \frac{\tan^{2}{\left(x \right)}}{x^{4}}\right)$$
Más detalles con x→-oo
Gráfico