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Răspuns:

-10

Explicație pas cu pas:

Sa pui si "n tinde la infinit" sub "lim", eu nu am mai pus:

[tex]a_n = n^2\cdot \left(\cos^5\dfrac{2}{n+1} - 1\right)\\\lim n^2\cdot \left(\cos^5\dfrac{2}{n+1} - 1\right) = \lim n^2 \cdot \left(\cos\dfrac{2}{n+1} - 1\right)\cdot \\ \left(\cos^4\dfrac{2}{n+1} + \cos^3\dfrac{2}{n+1} + \cos^2\dfrac{2}{n+1} + \cos\dfrac{2}{n+1} + 1\right)[/tex]

Limita din a doua paranteza este 5, deoarece cos⁴0 + cos³0 + cos²0 + cos 0 + 1 = 1 + 1 + 1 + 1 = 5.

Calculam limita din prima paranteza:

[tex]\lim n^2 \cdot \left(\cos\dfrac{2}{n+1} - 1\right) = \lim n^2 \cdot \left(- 2\cdot \sin^2\dfrac{1}{n+1}\right) = \\ = -2\cdot \lim \dfrac{\sin^2\frac{1}{n+1}}{\frac{1}{n^2}} = -2\lim \dfrac{\sin\frac{1}{n+1}}{\frac{1}{n}} \cdot \lim \dfrac{\sin\frac{1}{n+1}}{\frac{1}{n}}\\[/tex]

Cred ca mai mult ca sigur ai invatat limita remarcabila:

[tex]\boxed{\lim_{x\rightarrow 0} \dfrac{\sin x}{x} = 1}[/tex]

Se aplica si in cazul de fata, avand in vedere faptul ca:

[tex]\lim \dfrac{1}{n+1} = \lim \dfrac{1}{n} = 0[/tex]

Asadar vom avea:

[tex]-2\lim \dfrac{\sin\frac{1}{n+1}}{\frac{1}{n}} \cdot \lim \dfrac{\sin\frac{1}{n+1}}{\frac{1}{n}} = - 2\cdot 1 \cdot 1 = -2[/tex]

Atunci:

[tex]\lim a_n = (-2)\cdot 5 = \boxed{-10}[/tex]

Am folosit identitatile:

[tex]\bullet~a^n -b^n = (a - b)(a^{n-1} + a^{n-2}\cdot b + \ldots + a\cdot b^{n-2} + b^{n-1}),~\forall~n\in\mathbb{N}^\ast\\ \bullet \sin^2\frac{x}{2} = \dfrac{1 - \cos x}{2},~\forall~x\in\mathbb{R}\\\bullet \lim \dfrac{\sin x_n}{y_n} = 1,~daca~\lim x_n = \lim y_n = 0[/tex]