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(sin(x)^2-tan(x)^2)/x^4
Límite de la función/ tan(x)/ sin(x)/ (sin(x)^2-tan(x)^2)/x^4

Límite de la función (sin(x)^2-tan(x)^2)/x^4

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Solución

Ha introducido [src] 
     /   2         2   \
     |sin (x) - tan (x)|
 lim |-----------------|
x->0+|         4       |
     \        x        /
lim⁡x→0+(sin⁡2(x)−tan⁡2(x)x4)\lim_{x \to 0^+}\left(\frac{\sin^{2}{\left(x \right)} - \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
0/0,

tal que el límite para el numerador es
lim⁡x→0+(sin⁡2(x)−tan⁡2(x))=0\lim_{x \to 0^+}\left(\sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}\right) = 0
y el límite para el denominador es
lim⁡x→0+x4=0\lim_{x \to 0^+} x^{4} = 0
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
lim⁡x→0+(sin⁡2(x)−tan⁡2(x)x4)\lim_{x \to 0^+}\left(\frac{\sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}}{x^{4}}\right)
=
lim⁡x→0+(ddx(sin⁡2(x)−tan⁡2(x))ddxx4)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}\right)}{\frac{d}{d x} x^{4}}\right)
=
lim⁡x→0+(2sin⁡(x)cos⁡(x)−2tan⁡3(x)−2tan⁡(x)4x3)\lim_{x \to 0^+}\left(\frac{2 \sin{\left(x \right)} \cos{\left(x \right)} - 2 \tan^{3}{\left(x \right)} - 2 \tan{\left(x \right)}}{4 x^{3}}\right)
=
lim⁡x→0+(ddx(2sin⁡(x)cos⁡(x)−2tan⁡3(x)−2tan⁡(x))ddx4x3)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(2 \sin{\left(x \right)} \cos{\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→0+(−2sin⁡2(x)+2cos⁡2(x)−6tan⁡4(x)−8tan⁡2(x)−212x2)\lim_{x \to 0^+}\left(\frac{- 2 \sin^{2}{\left(x \right)} + 2 \cos^{2}{\left(x \right)} - 6 \tan^{4}{\left(x \right)} - 8 \tan^{2}{\left(x \right)} - 2}{12 x^{2}}\right)
=
lim⁡x→0+(ddx(−2sin⁡2(x)+2cos⁡2(x)−6tan⁡4(x)−8tan⁡2(x)−2)ddx12x2)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 2 \sin^{2}{\left(x \right)} + 2 \cos^{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→0+(−8sin⁡(x)cos⁡(x)−24tan⁡5(x)−40tan⁡3(x)−16tan⁡(x)24x)\lim_{x \to 0^+}\left(\frac{- 8 \sin{\left(x \right)} \cos{\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→0+(ddx(−8sin⁡(x)cos⁡(x)−24tan⁡5(x)−40tan⁡3(x)−16tan⁡(x))ddx24x)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- 8 \sin{\left(x \right)} \cos{\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→0+(sin⁡2(x)3−cos⁡2(x)3−5tan⁡6(x)−10tan⁡4(x)−17tan⁡2(x)3−23)\lim_{x \to 0^+}\left(\frac{\sin^{2}{\left(x \right)}}{3} - \frac{\cos^{2}{\left(x \right)}}{3} - 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→0+(sin⁡2(x)3−cos⁡2(x)3−5tan⁡6(x)−10tan⁡4(x)−17tan⁡2(x)3−23)\lim_{x \to 0^+}\left(\frac{\sin^{2}{\left(x \right)}}{3} - \frac{\cos^{2}{\left(x \right)}}{3} - 5 \tan^{6}{\left(x \right)} - 10 \tan^{4}{\left(x \right)} - \frac{17 \tan^{2}{\left(x \right)}}{3} - \frac{2}{3}\right)
=
−1-1
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 4 vez (veces)
Gráfica
02468-8-6-4-2-1010-10050
Trazar el gráfico
Otros límites con x→0, -oo, +oo, 1
lim⁡x→0−(sin⁡2(x)−tan⁡2(x)x4)=−1\lim_{x \to 0^-}\left(\frac{\sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}}{x^{4}}\right) = -1
Más detalles con x→0 a la izquierda
lim⁡x→0+(sin⁡2(x)−tan⁡2(x)x4)=−1\lim_{x \to 0^+}\left(\frac{\sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}}{x^{4}}\right) = -1
lim⁡x→∞(sin⁡2(x)−tan⁡2(x)x4)\lim_{x \to \infty}\left(\frac{\sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}}{x^{4}}\right)
Más detalles con x→oo
lim⁡x→1−(sin⁡2(x)−tan⁡2(x)x4)=−tan⁡2(1)+sin⁡2(1)\lim_{x \to 1^-}\left(\frac{\sin^{2}{\left(x \right)} - \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→1+(sin⁡2(x)−tan⁡2(x)x4)=−tan⁡2(1)+sin⁡2(1)\lim_{x \to 1^+}\left(\frac{\sin^{2}{\left(x \right)} - \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→−∞(sin⁡2(x)−tan⁡2(x)x4)\lim_{x \to -\infty}\left(\frac{\sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}}{x^{4}}\right)
Más detalles con x→-oo
Respuesta rápida [src] 
-1
−1-1
Abrir y simplificar
A la izquierda y a la derecha [src] 
     /   2         2   \
     |sin (x) - tan (x)|
 lim |-----------------|
x->0+|         4       |
     \        x        /
lim⁡x→0+(sin⁡2(x)−tan⁡2(x)x4)\lim_{x \to 0^+}\left(\frac{\sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}}{x^{4}}\right) 
-1
−1-1 
= -1.0
     /   2         2   \
     |sin (x) - tan (x)|
 lim |-----------------|
x->0-|         4       |
     \        x        /
lim⁡x→0−(sin⁡2(x)−tan⁡2(x)x4)\lim_{x \to 0^-}\left(\frac{\sin^{2}{\left(x \right)} - \tan^{2}{\left(x \right)}}{x^{4}}\right) 
-1
−1-1 
= -1.0
= -1.0
Respuesta numérica [src] 
-1.0
-1.0
Gráfico
Límite de la función (sin(x)^2-tan(x)^2)/x^4
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