Tính giới hạn \(\mathop {\lim }\limits_{x \to 1} \frac{{x + {x^2} + ... + {x^{2024}} - 2024}}{{x + {x^2} + {x^3} + {x^4} + {x^5} - 5}}\).

Chọn B

Trong mặt phẳng \(\left( {ABCD} \right)\)\(BM \cap AD = E.\)

Trong mặt phẳng \(\left( {SAD} \right)\)\(EN \cap SA = P \Rightarrow \left( {BMN} \right) \cap SA = P.\)

Ta có \(DA = DE,3NP = NE.\)

Trong tam giác \(SAD\) ta có \(\frac{{PS}}{{PA}}.\frac{{EA}}{{ED}}.\frac{{ND}}{{NS}} = 1 \Rightarrow \frac{{PS}}{{PA}} = \frac{1}{2}.\)

a) Ta có \(\mathop {\lim }\limits_{n \to + \infty } \left( {\sqrt {{n^2} + 5n} - n} \right) = \mathop {\lim }\limits_{n \to + \infty } \frac{{{n^2} + 5n - {n^2}}}{{\sqrt {{n^2} + 5n} + n}} = \mathop {\lim }\limits_{n \to + \infty } \frac{{5n}}{{\sqrt {{n^2} + 5n} + n}}\)

                                                =\(\mathop {\lim }\limits_{n \to + \infty } \frac{5}{{\sqrt {1 + \frac{5}{n}} + 1}} = \frac{5}{2}\).

b) Ta có \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {x + 6} \right) = 8 > 0\). \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {x - 2} \right) = 0\) và \(x - 2 < 0,\forall x \to {2^ - }\)

Vậy \(\mathop {\lim }\limits_{x \to {2^ - }} \frac{{x + 6}}{{x - 2}} = - \infty \)