Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\mathop {\lim }\limits_{x \to  - \infty } \left( {1 + x - 5{x^{15}}} \right) = \mathop {\lim }\limits_{x \to  - \infty } \left[ {{x^{15}}\left( { - 5 + \frac{1}{{{x^{14}}}} + \frac{1}{{{x^{15}}}}} \right)} \right]\\
\mathop {\lim }\limits_{x \to  - \infty } {x^{15}} =  - \infty \\
\mathop {\lim }\limits_{x \to  - \infty } \left( { - 5 + \frac{1}{{{x^{14}}}} + \frac{1}{{{x^{15}}}}} \right) =  - 5\\
 \Rightarrow \mathop {\lim }\limits_{x \to  - \infty } \left[ {{x^{15}}\left( { - 5 + \frac{1}{{{x^{14}}}} + \frac{1}{{{x^{15}}}}} \right)} \right] =  + \infty \\
b,\\
\mathop {\lim }\limits_{x \to  - \infty } \left( {x - \sqrt {{x^2} + 1} } \right) = \mathop {\lim }\limits_{x \to  - \infty } \left( {x - \sqrt {{x^2}\left( {1 + \frac{1}{{{x^2}}}} \right)} } \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \left( {x - \left| x \right|\sqrt {1 + \frac{1}{{{x^2}}}} } \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \left( {x + x\sqrt {1 + \frac{1}{{{x^2}}}} } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \to  - \infty  \Rightarrow x < 0 \Rightarrow \left| x \right| =  - x} \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \left[ {x\left( {1 + \sqrt {1 + \frac{1}{{{x^2}}}} } \right)} \right]\\
\mathop {\lim }\limits_{x \to  - \infty } x =  - \infty \\
\mathop {\lim }\limits_{x \to  - \infty } \left( {1 + \sqrt {1 + \frac{1}{{{x^2}}}} } \right) = 1 + \sqrt 1  = 2\\
 \Rightarrow \mathop {\lim }\limits_{x \to  - \infty } \left[ {x\left( {1 + \sqrt {1 + \frac{1}{{{x^2}}}} } \right)} \right] =  - \infty \\
c,\\
\mathop {\lim }\limits_{x \to  + \infty } \left( {x - \sqrt {{x^2} + 1} } \right) = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {x - \sqrt {{x^2} + 1} } \right)\left( {x + \sqrt {{x^2} + 1} } \right)}}{{x + \sqrt {{x^2} + 1} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2} - \left( {{x^2} + 1} \right)}}{{x + x\sqrt {1 + \frac{1}{{{x^2}}}} }} = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 1}}{{x + x\sqrt {1 + \frac{1}{{{x^2}}}} }} = \frac{{ - 1}}{{ + \infty }} = 0\\
\left( \begin{array}{l}
\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty \\
\mathop {\lim }\limits_{x \to  + \infty } \left( {1 + \sqrt {1 + \frac{1}{{{x^2}}}} } \right) = 1 + \sqrt 1  = 2\\
 \Rightarrow \mathop {\lim }\limits_{x \to  + \infty } \left[ {x\left( {1 + \sqrt {1 + \frac{1}{{{x^2}}}} } \right)} \right] =  + \infty 
\end{array} \right)\\
d,\\
\mathop {\lim }\limits_{x \to  + \infty } \left( {x - \sqrt {{x^2} + x + 2} } \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{\left( {x - \sqrt {{x^2} + x + 2} } \right)\left( {x + \sqrt {{x^2} + x + 2} } \right)}}{{x + \sqrt {{x^2} + x + 2} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{{x^2} - \left( {{x^2} + x + 2} \right)}}{{x + \sqrt {{x^2} + x + 2} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - \left( {x + 2} \right)}}{{x + \sqrt {{x^2} + x + 2} }}\\
 = \mathop {\lim }\limits_{x \to  + \infty } \frac{{ - 1 - \frac{2}{x}}}{{1 + \sqrt {1 + \frac{1}{x} + \frac{2}{{{x^2}}}} }} = \frac{{ - 1}}{{1 + \sqrt 1 }} =  - \frac{1}{2}\\
e,\\
\mathop {\lim }\limits_{x \to  - \infty } \left( {\sqrt[3]{{{x^3} + 4{x^2}}} + \sqrt {{x^2} - x + 1} } \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \left[ {\left( {\sqrt[3]{{{x^3} + 4{x^2}}} - x} \right) + \left( {\sqrt {{x^2} - x + 1}  + x} \right)} \right]\\
 = \mathop {\lim }\limits_{x \to  - \infty } \left[ {\frac{{\left( {{x^3} + 4{x^2}} \right) - {x^2}}}{{{{\sqrt[3]{{{x^3} + 4{x^2}}}}^2} + \sqrt[3]{{{x^3} + 4{x^2}}}.x + {x^2}}} + \frac{{\left( {{x^2} - x + 1} \right) - {x^2}}}{{\sqrt {{x^2} - x + 1}  - x}}} \right]\\
 = \mathop {\lim }\limits_{x \to  - \infty } \left[ {\frac{{4{x^2}}}{{{x^2}.{{\sqrt[3]{{1 + \frac{4}{x}}}}^2} + x.\sqrt[3]{{1 + \frac{4}{x}}}.x + {x^2}}} + \frac{{ - x + 1}}{{\left| x \right|.\sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}}  - x}}} \right]\\
 = \mathop {\lim }\limits_{x \to  - \infty } \left[ {\frac{4}{{{{\sqrt[3]{{1 + \frac{4}{x}}}}^2} + \sqrt[3]{{1 + \frac{4}{x}}} + 1}} + \frac{{ - x + 1}}{{ - x\sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}}  - x}}} \right]\,\,\,\,\left( {x \to  - \infty  \Rightarrow x < 0 \Rightarrow \left| x \right| =  - x} \right)\\
 = \mathop {\lim }\limits_{x \to  - \infty } \left[ {\frac{4}{{{{\sqrt[3]{{1 + \frac{4}{x}}}}^2} + \sqrt[3]{{1 + \frac{4}{x}}} + 1}} + \frac{{1 - \frac{1}{x}}}{{\sqrt {1 - \frac{1}{x} + \frac{1}{{{x^2}}}}  + 1}}} \right]\\
 = \frac{4}{3} + \frac{1}{2} = \frac{{11}}{6}
\end{array}\)

\(\begin{array}{l}
f,\\
\mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt[3]{{8{x^3} + x}} - \sqrt {{x^2} + 1} } \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } \left( {x.\sqrt[3]{{8 + \frac{1}{{{x^2}}}}} - x\sqrt {1 + \frac{1}{x}} } \right)\\
 = \mathop {\lim }\limits_{x \to  + \infty } \left[ {x.\left( {\sqrt[3]{{8 + \frac{1}{{{x^2}}}}} - \sqrt {1 + \frac{1}{x}} } \right)} \right]\\
\mathop {\lim }\limits_{x \to  + \infty } x =  + \infty \\
\mathop {\lim }\limits_{x \to  + \infty } \left[ {\sqrt[3]{{8 + \frac{1}{{{x^2}}}}} - \sqrt {1 + \frac{1}{x}} } \right] = \sqrt[3]{8} - \sqrt 1  = 1\\
 \Rightarrow \mathop {\lim }\limits_{x \to  + \infty } \left( {\sqrt[3]{{8{x^3} + x}} - \sqrt {{x^2} + 1} } \right) =  + \infty 
\end{array}\)