The given problem is:

$\lim_{{x \to 0}} \frac{x \cdot 3^x}{3^x - 1}$

To apply L'Hôpital's Rule, we first need to check if the limit results in an indeterminate form like $\frac{0}{0}$. Let’s check that:

1. As $x \to 0$, $3^x \to 1$.
2. The numerator $x \cdot 3^x \to 0$.
3. The denominator $3^x - 1 \to 0$.

Since we have the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule, which involves taking the derivative of the numerator and denominator.

Step 1: Derivatives

• Numerator: $f(x) = x \cdot 3^x$ $f'(x) = \frac{d}{dx}(x \cdot 3^x) = 3^x + x \cdot 3^x \ln(3)$
• Denominator: $g(x) = 3^x - 1$ $g'(x) = 3^x \ln(3)$

Step 2: Apply L'Hôpital's Rule

Now, we calculate the limit of the ratio of the derivatives:

$\lim_{{x \to 0}} \frac{3^x + x \cdot 3^x \ln(3)}{3^x \ln(3)}$

At $x = 0$:

• $3^x = 1$,
• The numerator becomes $1 + 0 = 1$,
• The denominator becomes $\ln(3)$.

Thus, the limit becomes:

$\frac{1}{\ln(3)}$

Final Answer:

$\lim_{{x \to 0}} \frac{x \cdot 3^x}{3^x - 1} = \frac{1}{\ln(3)}$

Would you like a detailed breakdown of any step, or do you have any additional questions?

5 Related Questions:

1. How does L'Hôpital's Rule work for other indeterminate forms, like $\infty / \infty$?
2. Can we solve this limit using a series expansion instead of L'Hôpital's Rule?
3. How would the limit change if the base of the exponential function was different, like $e^x$?
4. What is the derivative of the function $a^x$ in general, where $a$ is any positive constant?
5. Can you explain the conditions under which L'Hôpital's Rule is applicable?

Tip:

When applying L'Hôpital's Rule, make sure to check the indeterminate form before taking derivatives. Only certain forms like $0/0$ and $\infty/\infty$ allow the use of this rule.