Question

true or false: (f(x)) can be expressed as a single logarithmic function with behavior (lim_{x \to infty} f(x) = -infty) and (lim_{x \to 0^+} f(x) = infty).

true
false

Explanation:

⚡ Using what you learned: understanding exponential functions · Limits Involving Infinity and Asymptotes

Step 1: Analyze the standard logarithmic function

A single logarithmic function in its simplest form is:

$$ f(x) = \log_b(x) $$

For a standard base \( b > 1 \) (such as the natural logarithm \( \ln(x) \) where \( b = e \approx 2.718 \)):

• As \( x \to \infty \), \( \log_b(x) \to \infty \).
• As \( x \to 0^+ \), \( \log_b(x) \to -\infty \).

Step 2: Apply transformations to match the given limits

We want to find if there exists a single logarithmic function \( f(x) = a \log_b(x) \) or \( f(x) = \log_b(x) \) with a base \( 0 < b < 1 \) that satisfies:

1. \( \lim_{x \to \infty} f(x) = -\infty \)
2. \( \lim_{x \to 0^+} f(x) = \infty \)

Let's test a base \( b \) where \( 0 < b < 1 \) (for example, \( b = \frac{1}{2} \)):

$$ f(x) = \log_{1/2}(x) = -\log_2(x) $$

Now we evaluate the limits for this function:

• Limit as \( x \to \infty \):

$$ \lim_{x \to \infty} \log_{1/2}(x) = \lim_{x \to \infty} (-\log_2(x)) = -\infty $$

This matches the first condition.

• Limit as \( x \to 0^+ \):

$$ \lim_{x \to 0^+} \log_{1/2}(x) = \lim_{x \to 0^+} (-\log_2(x)) = -(-\infty) = \infty $$

This matches the second condition.

Since \( f(x) = \log_{1/2}(x) \) is a single logarithmic function that satisfies both conditions, the statement is true.

Answer:

True