To solve the limit [tex]\(\lim_{x \rightarrow 0} \frac{3^x - 1}{x}\)[/tex], we need to analyze the expression [tex]\(\frac{3^x - 1}{x}\)[/tex] as [tex]\(x\)[/tex] approaches 0. Here’s a step-by-step explanation:
1. Recognizing Indeterminate Form:
The given expression [tex]\(\frac{3^x - 1}{x}\)[/tex] as [tex]\(x\)[/tex] approaches 0 is an indeterminate form of the type [tex]\(\frac{0}{0}\)[/tex]. To evaluate such limits, we often use L'Hôpital's Rule.
2. Applying L'Hôpital's Rule:
L'Hôpital's Rule states that if [tex]\(\lim_{x \to c} \frac{f(x)}{g(x)}\)[/tex] is an indeterminate form [tex]\( \frac{0}{0} \)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], then:
[tex]\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}
\][/tex]
In our case:
[tex]\[
f(x) = 3^x - 1 \quad \text{and} \quad g(x) = x
\][/tex]
3. Differentiate Numerator and Denominator:
We differentiate the numerator and the denominator with respect to [tex]\(x\)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx}(3^x - 1) = 3^x \ln(3)
\][/tex]
[tex]\[
g'(x) = \frac{d}{dx}(x) = 1
\][/tex]
4. Compute the Limit of the Derivatives:
Now applying L'Hôpital's Rule:
[tex]\[
\lim_{x \to 0} \frac{3^x - 1}{x} = \lim_{x \to 0} \frac{3^x \ln(3)}{1}
\][/tex]
As [tex]\(x \to 0\)[/tex], [tex]\(3^x \to 3^0 = 1\)[/tex]:
[tex]\[
\lim_{x \to 0} 3^x \ln(3) = 1 \cdot \ln(3) = \ln(3)
\][/tex]
Therefore, the value of [tex]\(\lim_{x \rightarrow 0} \frac{3^x - 1}{x}\)[/tex] is:
[tex]\[
\boxed{\log 3}
\][/tex]
