Question

section: 2.11
which of the following pairs of statements correctly describes the end behavior of ( f(x)=2 ln x ) ?
(a) ( lim _{x
ightarrow 0^{+}} f(x)=-infty ) and ( lim _{x
ightarrow infty} f(x)=infty )
(b) ( lim _{x
ightarrow 0^{+}} f(x)=infty ) and ( lim _{x
ightarrow infty} f(x)=-infty )
(c) ( lim _{x
ightarrow-infty} f(x)=0^{+} ) and ( lim _{x
ightarrow infty} f(x)=infty )
(d) ( lim _{x
ightarrow-infty} f(x)=-infty ) and ( lim _{x
ightarrow infty} f(x)=0^{+} )

Explanation:

Step1: Analyze the domain of \(y = \ln x\)

The domain of \(y=\ln x\) is \(x>0\). So when considering the end - behavior, we only look at \(x\to0^{+}\) and \(x\to+\infty\) (since \(x =-\infty\) is not in the domain of \(f(x)=2\ln x\)).

Step2: Use the property of the natural - logarithm function

We know the property of the limit of the natural - logarithm function: \(\lim_{x\to0^{+}}\ln x=-\infty\) and \(\lim_{x\to+\infty}\ln x =+\infty\).

For the function \(f(x)=2\ln x\), by the constant - multiple rule of limits \(\lim_{x\to a}(k\cdot g(x))=k\cdot\lim_{x\to a}g(x)\) (\(k = 2\) is a constant and \(g(x)=\ln x\)):

\(\lim_{x\to0^{+}}f(x)=\lim_{x\to0^{+}}2\ln x=2\lim_{x\to0^{+}}\ln x\). Substituting \(\lim_{x\to0^{+}}\ln x =-\infty\), we get \(\lim_{x\to0^{+}}2\ln x=-\infty\).

\(\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}2\ln x=2\lim_{x\to+\infty}\ln x\). Substituting \(\lim_{x\to+\infty}\ln x=+\infty\), we get \(\lim_{x\to+\infty}2\ln x=+\infty\).

Answer:

A. \(\lim_{x\to0^{+}}f(x)=-\infty\) and \(\lim_{x\to+\infty}f(x)=\infty\)