Question

no calculator is allowed on this question.
which of the following statements are true for $f(x) = -2\log_{3}(x)$?
i. $\lim\limits_{x \to \infty} f(x) = \infty$
ii. $\lim\limits_{x \to 0^{+}} f(x) = \infty$
iii. the graph of $y = f(x)$ has a vertical asymptote at $x = 0$.
select one answer
a i only
b iii only
c ii and iii only
d i, ii, and iii

Explanation:

Step1: Analyze Statement I

For the function \( f(x) = -2\log_{3}(x) \), recall the behavior of the logarithmic function \( \log_{a}(x) \) (where \( a>1 \)). As \( x \to \infty \), \( \log_{3}(x) \to \infty \). Then \( -2\log_{3}(x) \to -\infty \) (since we multiply by -2). So \( \lim_{x \to \infty} f(x)=-\infty \), not \( \infty \). Thus, Statement I is false.

Step2: Analyze Statement II

As \( x \to 0^{+} \) (since the domain of \( \log_{3}(x) \) is \( x>0 \)), \( \log_{3}(x) \to -\infty \) (because for \( a > 1 \), \( \log_{a}(x) \) approaches \( -\infty \) as \( x \) approaches 0 from the right). Then \( -2\log_{3}(x)=-2\times(-\infty)=\infty \) (since multiplying a negative infinity by a negative number gives positive infinity). So \( \lim_{x \to 0^{+}} f(x)=\infty \), Statement II is true.

Step3: Analyze Statement III

A vertical asymptote of a function \( y = f(x) \) occurs where the function approaches \( \pm\infty \) as \( x \) approaches a value (from the left or right, within the domain). For \( f(x)=-2\log_{3}(x) \), the domain is \( x>0 \). As \( x \to 0^{+} \), \( f(x) \to \infty \) (from Statement II), so the graph has a vertical asymptote at \( x = 0 \). Thus, Statement III is true.

Answer:

C. II and III only