Question

find the limit. write ∞ or −∞ where appropriate.
\lim_{x \to 0^-} \frac{1}{3x}

\lim_{x \to 0^-} \frac{1}{3x} = \square (simplify your answer.)

Explanation:

Step1: Analyze the left - hand limit

We are finding the limit \(\lim_{x
ightarrow0^{-}}\frac{1}{3x}\). When \(x\) approaches \(0\) from the left (\(x
ightarrow0^{-}\)), \(x\) is a negative number that gets closer and closer to \(0\). Let's consider the behavior of the denominator \(3x\). If \(x\) is negative and approaching \(0\), then \(3x\) is also negative and approaching \(0\) (because multiplying a negative number by \(3\) still gives a negative number, and as \(x
ightarrow0\), \(3x
ightarrow0\)).

Step2: Analyze the behavior of the fraction

We have a fraction \(\frac{1}{3x}\). The numerator is \(1\) (a positive constant), and the denominator \(3x\) is negative and approaching \(0\). When we have a positive number divided by a negative number that is getting closer and closer to \(0\), the value of the fraction will approach \(-\infty\). For example, if we take \(x = - 0.1\), then \(\frac{1}{3x}=\frac{1}{3\times(-0.1)}=\frac{1}{-0.3}\approx - 3.33\). If we take \(x=-0.01\), then \(\frac{1}{3x}=\frac{1}{3\times(-0.01)}=\frac{1}{-0.03}\approx - 33.33\). As \(x\) gets closer to \(0\) from the left (i.e., \(x\) becomes a smaller negative number, like \(x = - 0.001\), \(\frac{1}{3x}=\frac{1}{3\times(-0.001)}=\frac{1}{-0.003}\approx - 333.33\)), the value of \(\frac{1}{3x}\) becomes more and more negative, approaching \(-\infty\).

Answer:

\(-\infty\)