Sr Examen
Otras calculadoras:
-
¿Cómo usar?
- Límite de la función:
-
Límite de tan(x)/x
-
Límite de sin(3*x)/x
-
Límite de (-16+x^2)/(-4+x)
-
Límite de (1-cos(x))/x
-
Expresiones idénticas
- cot(dos *x)/cot(ocho *x)
- cotangente de (2 multiplicar por x) dividir por cotangente de (8 multiplicar por x)
- cotangente de (dos multiplicar por x) dividir por cotangente de (ocho multiplicar por x)
- cot(2x)/cot(8x)
- cot2x/cot8x
- cot(2*x) dividir por cot(8*x)
-
Expresiones con funciones
- Cotangente cot
- cot(4*x)*tan(5*x)
- cot(2*x)
- cot(pi*x/2)/log(-2+x)
- cot(3*x)/cot(6*x)
- cot(x)*log(x)*sin(x)
- Cotangente cot
- cot(4*x)*tan(5*x)
- cot(2*x)
- cot(pi*x/2)/log(-2+x)
- cot(3*x)/cot(6*x)
- cot(x)*log(x)*sin(x)
Límite de la función cot(2*x)/cot(8*x)
Solución
Ha introducido
[src]
/cot(2*x)\ lim |--------| x->0+\cot(8*x)/
$$\lim_{x \to 0^+}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right)$$
Limit(cot(2*x)/cot(8*x), x, 0)
Método de l'Hopital
Tenemos la indeterminación de tipo
tal que el límite para el numerador es
$$\lim_{x \to 0^+} \frac{1}{\cot{\left(8 x \right)}} = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} \frac{1}{\cot{\left(2 x \right)}} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \frac{1}{\cot{\left(8 x \right)}}}{\frac{d}{d x} \frac{1}{\cot{\left(2 x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}}{\left(2 \cot^{2}{\left(2 x \right)} + 2\right) \cot^{2}{\left(8 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \frac{1}{2 \cot^{2}{\left(2 x \right)} + 2}}{\frac{d}{d x} \frac{\cot^{2}{\left(8 x \right)}}{8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{2 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)}}{\left(\frac{\left(- 16 \cot^{2}{\left(8 x \right)} - 16\right) \cot{\left(8 x \right)}}{8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}} + \frac{\left(- 8 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)} \cot^{2}{\left(8 x \right)} - 8 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)} - 8 \left(- 16 \cot^{2}{\left(8 x \right)} - 16\right) \cot^{2}{\left(2 x \right)} \cot{\left(8 x \right)}\right) \cot^{2}{\left(8 x \right)}}{\left(8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}\right)^{2}}\right) \left(2 \cot^{2}{\left(2 x \right)} + 2\right)^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{2 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)}}{\left(\frac{\left(- 16 \cot^{2}{\left(8 x \right)} - 16\right) \cot{\left(8 x \right)}}{8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}} + \frac{\left(- 8 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)} \cot^{2}{\left(8 x \right)} - 8 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)} - 8 \left(- 16 \cot^{2}{\left(8 x \right)} - 16\right) \cot^{2}{\left(2 x \right)} \cot{\left(8 x \right)}\right) \cot^{2}{\left(8 x \right)}}{\left(8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}\right)^{2}}\right) \left(2 \cot^{2}{\left(2 x \right)} + 2\right)^{2}}\right)$$
=
$$4$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+} \frac{1}{\cot{\left(8 x \right)}} = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+} \frac{1}{\cot{\left(2 x \right)}} = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \frac{1}{\cot{\left(8 x \right)}}}{\frac{d}{d x} \frac{1}{\cot{\left(2 x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}}{\left(2 \cot^{2}{\left(2 x \right)} + 2\right) \cot^{2}{\left(8 x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \frac{1}{2 \cot^{2}{\left(2 x \right)} + 2}}{\frac{d}{d x} \frac{\cot^{2}{\left(8 x \right)}}{8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{2 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)}}{\left(\frac{\left(- 16 \cot^{2}{\left(8 x \right)} - 16\right) \cot{\left(8 x \right)}}{8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}} + \frac{\left(- 8 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)} \cot^{2}{\left(8 x \right)} - 8 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)} - 8 \left(- 16 \cot^{2}{\left(8 x \right)} - 16\right) \cot^{2}{\left(2 x \right)} \cot{\left(8 x \right)}\right) \cot^{2}{\left(8 x \right)}}{\left(8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}\right)^{2}}\right) \left(2 \cot^{2}{\left(2 x \right)} + 2\right)^{2}}\right)$$
=
$$\lim_{x \to 0^+}\left(- \frac{2 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)}}{\left(\frac{\left(- 16 \cot^{2}{\left(8 x \right)} - 16\right) \cot{\left(8 x \right)}}{8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}} + \frac{\left(- 8 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)} \cot^{2}{\left(8 x \right)} - 8 \left(- 4 \cot^{2}{\left(2 x \right)} - 4\right) \cot{\left(2 x \right)} - 8 \left(- 16 \cot^{2}{\left(8 x \right)} - 16\right) \cot^{2}{\left(2 x \right)} \cot{\left(8 x \right)}\right) \cot^{2}{\left(8 x \right)}}{\left(8 \cot^{2}{\left(2 x \right)} \cot^{2}{\left(8 x \right)} + 8 \cot^{2}{\left(2 x \right)}\right)^{2}}\right) \left(2 \cot^{2}{\left(2 x \right)} + 2\right)^{2}}\right)$$
=
$$4$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 2 vez (veces)
Gráfica
A la izquierda y a la derecha
[src]
/cot(2*x)\ lim |--------| x->0+\cot(8*x)/
$$\lim_{x \to 0^+}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right)$$
4
$$4$$
= 4.0
/cot(2*x)\ lim |--------| x->0-\cot(8*x)/
$$\lim_{x \to 0^-}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right)$$
4
$$4$$
= 4.0
= 4.0
Otros límites con x→0, -oo, +oo, 1
$$\lim_{x \to 0^-}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right) = 4$$
Más detalles con x→0 a la izquierda
$$\lim_{x \to 0^+}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right) = 4$$
$$\lim_{x \to \infty}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right)$$
Más detalles con x→oo
$$\lim_{x \to 1^-}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right) = \frac{\tan{\left(8 \right)}}{\tan{\left(2 \right)}}$$
Más detalles con x→1 a la izquierda
$$\lim_{x \to 1^+}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right) = \frac{\tan{\left(8 \right)}}{\tan{\left(2 \right)}}$$
Más detalles con x→1 a la derecha
$$\lim_{x \to -\infty}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right)$$
Más detalles con x→-oo
Más detalles con x→0 a la izquierda
$$\lim_{x \to 0^+}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right) = 4$$
$$\lim_{x \to \infty}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right)$$
Más detalles con x→oo
$$\lim_{x \to 1^-}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right) = \frac{\tan{\left(8 \right)}}{\tan{\left(2 \right)}}$$
Más detalles con x→1 a la izquierda
$$\lim_{x \to 1^+}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right) = \frac{\tan{\left(8 \right)}}{\tan{\left(2 \right)}}$$
Más detalles con x→1 a la derecha
$$\lim_{x \to -\infty}\left(\frac{\cot{\left(2 x \right)}}{\cot{\left(8 x \right)}}\right)$$
Más detalles con x→-oo
Gráfico