a) Tìm cácgiới hạn sau i) \(\mathop {\lim }\limits_{x \to  - \infty } ( - 3{x^5} + 5{x^3} + x - 2)\)ii) \(\mathop {\lim }\limits_{

* Hướng dẫn giải

a) 

i) \(\mathop {\lim }\limits_{x \to  - \infty } ( - 3{x^5} + 5{x^3} + x - 2) = \mathop {\lim }\limits_{x \to  - \infty } {x^5}( - 3 + \frac{5}{{{x^2}}} + \frac{1}{{{x^4}}} - \frac{2}{{{x^5}}})\)

Mà \(\mathop {\lim }\limits_{x \to  - \infty } {x^5} =  - \infty ,\mathop {\lim }\limits_{x \to  - \infty } ( - 3 + \frac{5}{{{x^2}}} + \frac{1}{{{x^4}}} - \frac{2}{{{x^5}}}) =  - 3 < 0\)

Vậy \(\mathop {\lim }\limits_{x \to  - \infty } ( - 3{x^5} + 5{x^3} + x - 2) =  + \infty \)

ii) \(\mathop {\lim }\limits_{x \to  - \infty } \frac{{\sqrt {4{x^2} - 2x + 1}  - x}}{{2 - 3x}} = \mathop {\lim }\limits_{x \to  - \infty } \frac{{ - x\sqrt {4 - \frac{2}{x} + \frac{1}{{{x^2}}}}  - x}}{{2 - 3x}}\)

\(=\mathop {\lim }\limits_{x \to  - \infty } \frac{{ - \sqrt {4 - \frac{2}{x} + \frac{1}{{{x^2}}}}  - 1}}{{\frac{2}{x} - 3}}=1\)

b) \(y = {\left( {m + \frac{n}{{{x^2}}}} \right)^4} \Rightarrow y' = 4{\left( {m + \frac{n}{{{x^2}}}} \right)^3}\left( {m + \frac{n}{{{x^2}}}} \right)' = 4{\left( {m + \frac{n}{{{x^2}}}} \right)^3}\left( { - \frac{{2n}}{{{x^3}}}} \right) =  - \frac{{8n}}{{{x^3}}}{\left( {m + \frac{n}{{{x^2}}}} \right)^3}\)

Vậy \(y'(1) =  - 8n{\left( {m + n} \right)^3}\)