Question

Formulate a precise definition of

\(\mathop {{\rm{lim}}}\limits_{x \to - \infty } f\left( x \right) = - \infty \)

Then use your definition to prove that

\(\mathop {{\rm{lim}}}\limits_{x \to - \infty } \left( {1 + {x^3}} \right) = - \infty \)

Short answer

• Assume that \(f\) be a function defined on the interval \(\left( { - \infty ,a} \right)\). Then \(\mathop {{\rm{lim}}}\limits_{x \to - \infty } f\left( x \right) = - \infty \) which implies that for every \(M < 0\) there exist a negative number \(N\) such that \(f\left( x \right) < M\) for every \(x < N\).
• It is proved that \(\mathop {{\rm{lim}}}\limits_{x \to - \infty } \left( {1 + {x^3}} \right) = - \infty \).

Step-by-step solution

Write precise definition

Assume that \(f\) be a function defined on the interval \(\left( { - \infty ,a} \right)\). Then \(\mathop {{\rm{lim}}}\limits_{x \to - \infty } f\left( x \right) = - \infty \)which implies that for every \(M < 0\) there exist a negative number \(N\) such that \(f\left( x \right) < M\) for every \(x < N\).

use precise definition to prove

Given\(M < 0\), we have to find the negative number\(N\)such that\(1 + {x^3} < M\).

Simplify the condition\(1 + {x^3} < M\)for\(x\).

\(\begin{array}{c}1 + {x^3} < M\\{x^3} < M - 1\\x < \sqrt(3){{M - 1}}\end{array}\)

Assume\(N = \sqrt(3){{M - 1}}\), such that\(x < \sqrt(3){{M - 1}}\). This implies that\(1 + {x^3} < M\). By using the definition, we have\(\mathop {{\rm{lim}}}\limits_{x \to - \infty } \left( {1 + {x^3}} \right) = - \infty \).

Thus, it is proved that \(\mathop {{\rm{lim}}}\limits_{x \to - \infty } \left( {1 + {x^3}} \right) = - \infty \).