Evaluate the following limits :
`Lim_( x to 0) (3^(2x)-1)/(2^(3x)-1)`

Answer

To evaluate the limit

\[ L = \lim_{x \to 0} \frac{3^{2x} - 1}{2^{3x} - 1}, \]

we notice that substituting \(x = 0\) gives us the indeterminate form \(\frac{0}{0}\). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), we can take the derivative of the numerator and the denominator.

**Step 1: Differentiate the numerator and denominator.**

The numerator is \(f(x) = 3^{2x} - 1\). The derivative of \(f(x)\) is:

\[ f'(x) = \frac{d}{dx}(3^{2x}) = 3^{2x} \cdot \ln(3) \cdot 2 = 2 \cdot 3^{2x} \ln(3). \]

The denominator is \(g(x) = 2^{3x} - 1\). The derivative of \(g(x)\) is:

\[ g'(x) = \frac{d}{dx}(2^{3x}) = 2^{3x} \cdot \ln(2) \cdot 3 = 3 \cdot 2^{3x} \ln(2). \]

**Step 2: Apply L'Hôpital's Rule.**

Now we can apply L'Hôpital's Rule:

\[ L = \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{2 \cdot 3^{2x} \ln(3)}{3 \cdot 2^{3x} \ln(2)}. \]

**Step 3: Substitute \(x = 0\) into the derivatives.**

Now we substitute \(x = 0\):

\[ L = \frac{2 \cdot 3^{0} \ln(3)}{3 \cdot 2^{0} \ln(2)} = \frac{2 \cdot 1 \cdot \ln(3)}{3 \cdot 1 \cdot \ln(2)} = \frac{2 \ln(3)}{3 \ln(2)}. \]

Thus, the limit evaluates to:

\[ L = \frac{2 \ln(3)}{3 \ln(2)}. \]