PROBLEMA 5 DE EVAU MATEMATICAS II ARAGÓN

Calcule lo siguient limite:

$$\displaystyle\lim_{x\rightarrow0^+}\big((1+x-\text{sen}(x))^{1/x^3}\big)$$

$$\displaystyle\lim_{x\rightarrow0^+}\big((1+x-\text{sen}(x))^{1/x^3}\big)=1^{\infty}$$
Clamamos $L$ a lo limite que nos piden. Aplicando logaritmos:

$$\displaystyle\begin{align} \ln L=\ln\lim_{x\rightarrow0^+}\big((1+x-\text{sen}(x))^{1/x^3}\big)=\lim_{x\rightarrow0^+}\ln\big((1+x-\text{sen}(x))^{1/x^3}\big)=\\=\lim_{x\rightarrow0^+}\dfrac1{x^3}\cdot\ln(1+x-\text{sen}(x))=\lim_{x\rightarrow0^+}\dfrac{\ln(1+x-\text{sen}(x))}{x^3}=\dfrac00\end{align}$$
Resolvemos esta indeterminación aplicando la regla de L’Hôpital:

$$\displaystyle\begin{align}\ln L=\lim_{x\rightarrow0^+}\dfrac{\ln(1+x-\text{sen}(x))}{x^3}\underset{L’H}=\lim_{x\rightarrow0^+}\dfrac{\frac{1-\cos(x)}{1+x-\text{sen}(x)}}{3x^2}=\\=\lim_{x\rightarrow0^+}\dfrac{1-\cos(x)}{3x^2+3x^3-3x^2\text{sen}(x)}=\dfrac00=\\\underset{L’H}=\lim_{x\rightarrow0^+}\dfrac{\text{sen}(x)}{6x+9x^2-6x\text{sen}(x)-3x^2\cos(x)}=\dfrac00=\\\underset{L’H}=\lim_{x\rightarrow0^+}\dfrac{\cos(x)}{6+18x-6\text{sen}(x)-6x\cos(x)-6x\cos(x)+3x^2\text{sen}(x)}=\dfrac16\end{align}$$
Dimpués, si $\ln L=\frac16$ alavez:

$$\boxed{L=y^{1/6}}$$