$\lim_{x \to 0} \frac{\ln(1 + "arctg"(x))^x}{e - e^{\cos^4(x)}}$

A cura di: Gianni Sammito

Calcolare

 

 $lim_{x to 0} frac{ln(1 + "arctg"(x))^x}{e – e^{cos^4(x)}}$

 

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Rciordando le proprietà dei logaritmi, e raccogliendo al denominatore un fattore $-e$, si ottiene

 

 

$lim_{x to 0} frac{x}{-e} frac{ln(1 + "arctg"(x))}{e^{cos^4(x) – 1} – 1} = lim_{x to 0} frac{x}{-e} frac{ln(1 + "arctg"(x))}{"arctg"(x)} frac{"arctg"(x)}{cos^4(x) – 1} frac{cos^4(x) – 1}{e^{cos^4(x) – 1} – 1} =$

$ = lim_{x to 0} frac{x}{-e} frac{ln(1 + "arctg"(x))}{"arctg"(x)} frac{"arctg"(x)}{(cos^2(x) – 1)(cos^2(x) + 1)} frac{cos^4(x) – 1}{e^{cos^4(x) – 1} – 1} =$

 $ = lim_{x to 0} frac{x}{e} frac{ln(1 + "arctg"(x))}{"arctg"(x)} frac{"arctg"(x)}{(1 – cos^2(x))(cos^2(x) + 1)} frac{cos^4(x) – 1}{e^{cos^4(x) – 1} – 1} =$

$ = lim_{x to 0} frac{1}{e (cos^2(x) + 1)} frac{ln(1 + "arctg"(x))}{"arctg"(x)} frac{x "arctg"(x)}{sin^2(x)} frac{cos^4(x) – 1}{e^{cos^4(x) – 1} – 1} =$

$ = lim_{x to 0} frac{1}{e (cos^2(x) + 1)} frac{ln(1 + "arctg"(x))}{"arctg"(x)} frac{x}{sin(x)} frac{"arctg"(x)}{x} frac{x}{sin(x)} frac{cos^4(x) – 1}{e^{cos^4(x) – 1} – 1}$

 

Ricordando i limiti notevoli

 

$lim_{t to 0} frac{e^t – 1}{t} =1$

 

$lim_{t to  0} frac{ln(1 + t)}{t} = 1$

 

$lim_{t to 0} frac{"arctg"(t)}{t} = 1$

 

$lim_{t to 0} frac{sin(t)}{t} = 1$

 

e osservando che $"arctg"(x) to 0$ e $cos^4(x) – 1 to 0$ per $x to 0$, si ottiene

 

$ lim_{x to 0} frac{1}{e (cos^2(x) + 1)} frac{ln(1 + "arctg"(x))}{"arctg"(x)} frac{x}{sin(x)} frac{"arctg"(x)}{x} frac{x}{sin(x)} frac{cos^4(x) – 1}{e^{cos^4(x) – 1} – 1} = frac{1}{2e} cdot 1 cdot 1 cdot 1 cdot 1 cdot 1 = frac{1}{2e}$

 

FINE