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Rayzen
a fost răspuns

Calculați limita:

[tex]\lim\limits_{x\rightarrow 0} \dfrac{x^3 - \text{tg}^3 x}{x^5}[/tex]

Răspuns :

Am gasit o demonstratie faina (bine de fapta n-am gasit-o eu,m-a ajutat altcineva ) .Impartim in doua limite:[tex]\displaystyle\limit\lim_{x\to 0} \left(\dfrac{x^3-tg ^3 x}{x^5}\right)= \displaystyle\limit\lim_{x\to 0} \left(\dfrac{(x-tg\ x)}{x^3}\cdot \dfrac{ x^2+tg\ x\cdot x+tg^2\ x}{x^2}\right)=\\ =\displaystyle\limit\lim_{x\to 0} \left(\dfrac{(x-tg\ x)}{x^3}\right) \cdot \displaystyle\limit\lim_{x\to 0} \left( \dfrac{ x^2+tg\ x\cdot x+tg^2\ x}{x^2}\right) [/tex]
Calculam limitele separat:
i) Pentru prima aplicand de trei ori l'Hopital se obtine :
[tex]\displaystyle\limit\lim_{x\to 0} \left(\dfrac{-1}{3\cdot \cos^2 x(\cos^2 x-x\cdot \sin 2x)}\right)= \boxed{\dfrac{-1}{3}}[/tex]

i)Pentru a doua se obtine:
[tex]\displaystyle\limit\lim_{x\to 0} \left(\dfrac{x^2+tg\ x\cdot x+tg^2 x}{x^2}\right)= \displaystyle\limit\lim_{x\to 0} \left(1+\dfrac{tg\ x}{x}+\dfrac{tg ^2\ x}{x^2} \right)=3\\ [/tex]
Deci limita  este:
[tex]3\cdot \left(-\dfrac{1}{3}\right) =\boxed{-1}[/tex]

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