Question

42: Prove, using Definition 6, that \(\mathop {{\rm{lim}}}\limits_{x \to - 3} \frac{1}{{{{\left( {x + 3} \right)}^4}}} = \infty \).

Short answer

It is proved that \(\mathop {{\rm{lim}}}\limits_{x \to - 3} \frac{1}{{{{\left( {x + 3} \right)}^4}}} = \infty \).

Step-by-step solution

Precise Definition of an infinite limit

According to definition 6, any function \(y\) defined on an open interval contains the number \(a\) and function at \(a\) itself. Then \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = \infty \) which implies that for every \(M > 0\) there is a positive number \(\delta > 0\) such that if \(0 < \left| {x - a} \right| < \delta \) implies that \(f\left( x \right) > M\).

Apply definition 6. We have for every \(M > 0\) there exist \(\delta > 0\)such that \(0 < \left| {x + 3} \right| < \delta \) implies that \(\frac{1}{{{{\left( {x + 3} \right)}^4}}} > M\).

Simplify the obtained condition

Rewrite and Take 4^th root on both sides of the obtained inequality.

\(\begin{array}{c}\frac{1}{{{{\left( {x + 3} \right)}^4}}} > M\\{\left( {x + 3} \right)^4} < \frac{1}{M}\\\left| {x + 3} \right| < \frac{1}{{\sqrt(4){M}}}\end{array}\)

This implies that\(\delta = \frac{1}{{\sqrt(4){M}}}\). Therefore, \(0 < \left| {x + 3} \right| < \frac{1}{{\sqrt(4){M}}}\) and \(\frac{1}{{{{\left( {x + 3} \right)}^4}}} > M\) which implies that \(\frac{1}{{{{\left( {x + 3} \right)}^4}}}\) tends to \(\frac{1}{{{\infty ^4}}}\) when \(x\) to \( - 3\) , that is, \(\mathop {{\rm{lim}}}\limits_{x \to - 3} \frac{1}{{{{\left( {x + 3} \right)}^4}}} = \infty \).