Càlcul del límit

Es demana el valor del límit: $$\lim_{x \to 0} \frac{\exp(\sin^2(x)) – 1 – x^2}{x^4}.$$

Pas 1: Desenvolupament de (\sin(x))

$$\sin(x) = x – \frac{x^3}{6} + O(x^5).$$

Pas 2: $\sin^2(x)$

$$\sin^2(x) = \left(x – \frac{x^3}{6} + O(x^5)\right)^2 = x^2 – 2 \cdot x \cdot \frac{x^3}{6} + O(x^4) = x^2 – \frac{x^4}{3} + O(x^4).$$
Més precisament:
$$\sin^2(x) = x^2 – \frac{x^4}{3} + O(x^6).$$

Pas 3: Desenvolupament de $\exp(u)$ per $u = \sin^2(x)$

Sabem que:
$$\exp(u) = 1 + u + \frac{u^2}{2} + \frac{u^3}{6} + O(u^4).$$
Substituïm $u = \sin^2(x) = x^2 – \frac{x^4}{3} + O(x^6)$:

• $u = x^2 – \frac{x^4}{3} + O(x^6)$,
• $u^2 = (x^2)^2 + O(x^6) = x^4 + O(x^6)$,
• $u^3 = O(x^6)), (u^4 = O(x^8)$.

Aleshores:
$$\exp(u) = 1 + \left(x^2 – \frac{x^4}{3} + O(x^6)\right) + \frac{1}{2}(x^4 + O(x^6)) + \frac{1}{6} O(x^6) + O(x^8).$$
$$= 1 + x^2 – \frac{x^4}{3} + \frac{x^4}{2} + O(x^6) = 1 + x^2 + \left(\frac{1}{2} – \frac{1}{3}\right)x^4 + O(x^6) = 1 + x^2 + \frac{1}{6} x^4 + O(x^6).$$

Pas 4: Numerador

$$\exp(\sin^2(x)) – 1 – x^2 = \left(1 + x^2 + \frac{1}{6} x^4 + O(x^6)\right) – 1 – x^2 = \frac{1}{6} x^4 + O(x^6).$$

Pas 5: Límit

$$\frac{\exp(\sin^2(x)) – 1 – x^2}{x^4} = \frac{\frac{1}{6} x^4 + O(x^6)}{x^4} = \frac{1}{6} + O(x^2).$$
Prenent $x \to 0$:
$$\lim_{x \to 0} \frac{\exp(\sin^2(x)) – 1 – x^2}{x^4} = \frac{1}{6}.$$

Resposta final

El valor del límit és:
$$\boxed{\dfrac{1}{6}}$$