Question

44: Suppose that \(\mathop {{\rm{lim}}}\limits_{x \to a} f\left( x \right) = \infty \) and \(\mathop {{\rm{lim}}}\limits_{x \to a} g\left( x \right) = c\), where \(c\) is a real number. Prove each statement.

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

(b) \(\mathop {{\rm{lim}}}\limits_{x \to a} \left( {f\left( x \right)g\left( x \right)} \right) = \infty \)if \(c > 0\)

(c) \(\mathop {{\rm{lim}}}\limits_{x \to a} \left( {f\left( x \right)g\left( x \right)} \right) = - \infty \)if \(c < 0\)

Short answer

(a) It is proved that \(\mathop {{\rm{lim}}}\limits_{x \to a} \left( {f\left( x \right) + g\left( x \right)} \right) = \infty \).

(b) It is proved that \(\mathop {{\rm{lim}}}\limits_{x \to a} \left( {f\left( x \right)g\left( x \right)} \right) = \infty \) if \(c > 0\).

(c) It is proved that \(\mathop {{\rm{lim}}}\limits_{x \to a} \left( {f\left( x \right) + g\left( x \right)} \right) = - \infty \) if \(c < 0\).

Step-by-step solution

(a) Step 1: Precise Definition oflimit

Using the definition of limit for a function \(f\left( x \right)\), consider \(M\)be a given number, and \(\mathop {{\rm{lim}}}\limits_{x \to a} f\left( x \right) = \infty \)there exists\({\delta _1} > 0\) such that \(0 < \left| {x - a} \right| < {\delta _1}\)which implies that \(f\left( x \right) > M + 1 - c\).

Using the definition of limit for a function \(g\left( x \right)\), consider \(M\) be a given number and \(\mathop {{\rm{lim}}}\limits_{x \to a} g\left( x \right) = c\) there exist \({\delta _2} > 0\) such that \(0 < \left| {x - a} \right| < {\delta _2}\) which implies that \(\left| {g\left( x \right) - c} \right| < 1\), that is, \(g\left( x \right) > c - 1\).

Writethe condition for \(f\left( x \right) + g\left( x \right)\)

Assume that \(\delta \) is the minimum of \({\delta _1}\) and \({\delta _2}\), that is, \(\delta = {\rm{min}}\left\{ {{\delta _1},{\delta _2}} \right\}\). Then this also implies that \(\left| {x - a} \right| < \delta \).

Simplify the condition for \(f\left( x \right) + g\left( x \right)\).

\(\begin{array}{c}f\left( x \right) + g\left( x \right) > M + 1 - c + c - 1\\f\left( x \right) + g\left( x \right) > M\end{array}\)

Precise Definition of an infinite limit

According to definition 6, for any function \(f\) defined on open interval which 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 \(f\left( x \right) > M\).

This implies for \(0 < \left| {x - a} \right| < \delta \) and \(f\left( x \right) + g\left( x \right) > M\) which implies that \(f\left( x \right) + g\left( x \right)\) tends to \(\infty \) when \(x\) to \(a\), that is, \(\mathop {{\rm{lim}}}\limits_{x \to a} \left( {f\left( x \right) + g\left( x \right)} \right) = \infty \).

(b) Step 4: Precise Definition of limit

Using the definition of limit for a function \(g\left( x \right)\), consider \(M > 0\) and \(\mathop {{\rm{lim}}}\limits_{x \to a} g\left( x \right) = c > 0\) there exist \({\delta _1} > 0\) such that \(0 < \left| {x - a} \right| < {\delta _1}\) which implies that \(\left| {g\left( x \right) - c} \right| < \frac{c}{2}\), that is, \(g\left( x \right) > \frac{c}{2}\).

Using the definition of limit for a function \(f\left( x \right)\), consider \(M\) be a given number, and \(\mathop {{\rm{lim}}}\limits_{x \to a} f\left( x \right) = \infty \) there exist \({\delta _2} > 0\) such that \(0 < \left| {x - a} \right| < {\delta _2}\) which implies that \(f\left( x \right) > \frac{{2M}}{c}\).

Write the condition for \(f\left( x \right)g\left( x \right)\)

Assume that \(\delta \) is the minimum of \({\delta _1}\) and \({\delta _2}\), that is, \(\delta = {\rm{min}}\left\{ {{\delta _1},{\delta _2}} \right\}\). Then this also implies that \(\left| {x - a} \right| < \delta \).

Simplify the condition for \(f\left( x \right)g\left( x \right)\).

\(\begin{array}{c}f\left( x \right)g\left( x \right) > \left( {\frac{{2M}}{c}} \right)\left( {\frac{c}{2}} \right)\\f\left( x \right)g\left( x \right) > M\end{array}\)

This implies for \(0 < \left| {x - a} \right| < \delta \) and \(f\left( x \right)g\left( x \right) > M\) which implies that \(f\left( x \right)g\left( x \right)\) tends to \(\infty \) when \(x\) to \(a\), that is, \(\mathop {{\rm{lim}}}\limits_{x \to a} \left( {f\left( x \right)g\left( x \right)} \right) = \infty \) if \(c > 0\).

(c) Step 6: Precise Definition of limit

Using definition of limit for function\(g\left( x \right)\), consider \(N < 0\) and \(\mathop {{\rm{lim}}}\limits_{x \to a} g\left( x \right) = c < 0\) there exist \({\delta _1} > 0\) such that \(0 < \left| {x - a} \right| < {\delta _1}\) which implies that \(\left| {g\left( x \right) - c} \right| < - \frac{c}{2}\), that is, \(g\left( x \right) < \frac{c}{2}\).

Using definition of limit for function \(f\left( x \right)\), consider \(N\) be a given number and \(\mathop {{\rm{lim}}}\limits_{x \to a} f\left( x \right) = \infty \) there exist \({\delta _2} > 0\) such that \(0 < \left| {x - a} \right| < {\delta _2}\) which implies that \(f\left( x \right) > \frac{{2N}}{c}\).

Write the condition for \(f\left( x \right)g\left( x \right)\)

Assume that \(\delta \) is the minimum of \({\delta _1}\) and \({\delta _2}\), that is, \(\delta = {\rm{min}}\left\{ {{\delta _1},{\delta _2}} \right\}\). Then this also implies that \(\left| {x - a} \right| < \delta \).

Simplify the condition for \(f\left( x \right)g\left( x \right)\).

\(\begin{array}{c}f\left( x \right)g\left( x \right) < \left( {\frac{{2N}}{c}} \right)\left( {\frac{c}{2}} \right)\\f\left( x \right)g\left( x \right) < N\end{array}\)

This implies for \(0 < \left| {x - a} \right| < \delta \) and \(f\left( x \right)g\left( x \right) < N\) which implies that \(f\left( x \right)g\left( x \right)\) tends to \( - \infty \) when \(x\) to \(a\), that is, \(\mathop {{\rm{lim}}}\limits_{x \to a} \left( {f\left( x \right)g\left( x \right)} \right) = - \infty \) if \(c < 0\).