Question

When $a \gt 0$, $\lim_{x \to 0} \frac{a^x - 1}{x} =$

• A. $\log a$
• B. $0$
• C. $\log (x+1)$
• D. $\log (x \cdot a)$

Hint:

This limit is actually the definition of the derivative of $a^x$ at $x=0$. Since $\frac{d}{dx}(a^x) = a^x \ln a$, at $x=0$ it becomes $a^0 \ln a = \ln a$.

Answer 1

The Correct Option is A

This is a standard limit that can be evaluated using L'Hôpital's Rule since it results in an indeterminate form $0/0$. 1. Check Indeterminate Form: Substitute $x = 0$: $$\frac{a^0 - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}$$

2. Apply L'Hôpital's Rule: Differentiate the numerator and the denominator with respect to $x$: $$\frac{d}{dx}(a^x - 1) = a^x \log a$$ $$\frac{d}{dx}(x) = 1\lt strong\gt 3. Evaluate the Limit:\lt /strong\gt \lim_{x \to 0} \frac{a^x \log a}{1} = a^0 \log a$$ Since $a^0 = 1$: $$\text{Limit} = 1 \cdot \log a = \log a$$ The base of the logarithm is $e$ (natural logarithm), often written as $\ln a$.