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[math]\lim_{x \to 0} \frac{ln(1 + \text{arctg}(x))^x}{e - e^{\\cos^4(x)}}[/math]


Rciordando le proprietà  dei logaritmi, e raccogliendo al denominatore un fattore
[math]-e[/math]
, si ottiene

[math]\lim_{x \to 0} \frac{x}{-e} \frac{ln(1 + \text{arctg}(x))}{e^{\\cos^4(x) - 1} - 1} = \lim_{x \to 0} \frac{x}{-e} \frac{ln(1 + \text{arctg}(x))}{\text{arctg}(x)} \frac{\text{arctg}(x)}{\\cos^4(x) - 1} \frac{\\cos^4(x) - 1}{e^{\\cos^4(x) - 1} - 1} =[/math]

[math] = \lim_{x \to 0} \frac{x}{-e} \frac{ln(1 + \text{arctg}(x))}{\text{arctg}(x)} \frac{\text{arctg}(x)}{(\\cos^2(x) - 1)(\\cos^2(x) + 1)} \frac{\\cos^4(x) - 1}{e^{\\cos^4(x) - 1} - 1} =[/math]

[math] = \lim_{x \to 0} \frac{x}{e} \frac{ln(1 + \text{arctg}(x))}{\text{arctg}(x)} \frac{\text{arctg}(x)}{(1 - \\cos^2(x))(\\cos^2(x) + 1)} \frac{\\cos^4(x) - 1}{e^{\\cos^4(x) - 1} - 1} =[/math]

[math] = \lim_{x \to 0} \frac{1}{e (\\cos^2(x) + 1)} \frac{ln(1 + \text{arctg}(x))}{\text{arctg}(x)} \frac{x \text{arctg}(x)}{\\sin^2(x)} \frac{\\cos^4(x) - 1}{e^{\\cos^4(x) - 1} - 1} =[/math]

[math] = \lim_{x \to 0} \frac{1}{e (\\cos^2(x) + 1)} \frac{ln(1 + \text{arctg}(x))}{\text{arctg}(x)} \frac{x}{\\sin(x)} \frac{\text{arctg}(x)}{x} \frac{x}{\\sin(x)} \frac{\\cos^4(x) - 1}{e^{\\cos^4(x) - 1} - 1}[/math]

Ricordando i limiti notevoli

[math]\lim_{t \to 0} \frac{e^t - 1}{t} =1[/math]

[math]\lim_{t \to 0} \frac{ln(1 + t)}{t} = 1[/math]

[math]\lim_{t \to 0} \frac{\text{arctg}(t)}{t} = 1[/math]

[math]\lim_{t \to 0} \frac{\\sin(t)}{t} = 1[/math]

e osservando che

[math]\text{arctg}(x) \to 0[/math]
e
[math]\\cos^4(x) - 1 \to 0[/math]
per
[math]x \to 0[/math]
, si ottiene

[math] \lim_{x \to 0} \frac{1}{e (\\cos^2(x) + 1)} \frac{ln(1 + \text{arctg}(x))}{\text{arctg}(x)} \frac{x}{\\sin(x)} \frac{\text{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}[/math]

FINE

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