Tính giới hạn \(\mathop {\lim }\limits_{x \to  - \infty } \left( {{{\left| x \right|}^5} + x + 1} \right)\)

\(\mathop {\lim }\limits_{x \to  - \infty } \left( {\left| {{x^5}} \right| + x + 1} \right) = \mathop {\lim }\limits_{x \to  - \infty } \left| {{x^5}} \right|\left( {1 + \frac{x}{{\left| {{x^5}} \right|}} + \frac{1}{{\left| {{x^5}} \right|}}} \right)\).

Xét giới hạn tại \( - \infty \) nên \(x < 0\) hay \({x^5} < 0\). Từ đó suy ra \(\left| {{x^5}} \right| =  - {x^5}\).

Ta có:

\(\mathop {\lim }\limits_{x \to  - \infty } \left| {{x^5}} \right| = \mathop {\lim }\limits_{x \to  - \infty } ( - {x^5}) =  - \mathop {\lim }\limits_{x \to  - \infty } {x^5} =  + \infty \);

\(\mathop {\lim }\limits_{x \to  - \infty } \left( {1 + \frac{x}{{\left| {{x^5}} \right|}} + \frac{1}{{\left| {{x^5}} \right|}}} \right) = \mathop {\lim }\limits_{x \to  - \infty } \left( {1 + \frac{x}{{ - {x^5}}} + \frac{1}{{ - {x^5}}}} \right) = \mathop {\lim }\limits_{x \to  - \infty } \left( {1 - \frac{1}{{{x^4}}} - \frac{1}{{{x^5}}}} \right) = 1 - 0 - 0 = 1\).

Suy ra \(\mathop {\lim }\limits_{x \to  - \infty } \left| {{x^5}} \right|\left( {1 + \frac{x}{{\left| {{x^5}} \right|}} + \frac{1}{{\left| {{x^5}} \right|}}} \right) =  + \infty \).