`lim_(xrarr oo) (4^(1//n)-1)/(3^(1//n)-1)` is equal to

Options

A. `log_(4)3`
B. `log_(3)4`
C. `1`
D. none of these

Explanation

To solve the limit \( \lim_{n \to \infty} \frac{4^{1/n} - 1}{3^{1/n} - 1} \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{n \to \infty} \frac{4^{1/n} - 1}{3^{1/n} - 1} \]
As \( n \to \infty \), \( \frac{1}{n} \to 0 \). Therefore, both \( 4^{1/n} \) and \( 3^{1/n} \) approach 1, leading to the indeterminate form \( \frac{0}{0} \).

### Step 2: Substitute \( a = \frac{1}{n} \)
Let \( a = \frac{1}{n} \). As \( n \to \infty \), \( a \to 0 \). We can rewrite the limit as:
\[ \lim_{a \to 0} \frac{4^a - 1}{3^a - 1} \]

### Step 3: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We differentiate the numerator and the denominator with respect to \( a \):
- The derivative of the numerator \( 4^a - 1 \) is \( 4^a \ln(4) \).
- The derivative of the denominator \( 3^a - 1 \) is \( 3^a \ln(3) \).

Thus, we have:
\[ \lim_{a \to 0} \frac{4^a \ln(4)}{3^a \ln(3)} \]

### Step 4: Evaluate the limit
As \( a \to 0 \), both \( 4^a \) and \( 3^a \) approach 1. Therefore, we can simplify the limit:
\[ \lim_{a \to 0} \frac{4^a \ln(4)}{3^a \ln(3)} = \frac{1 \cdot \ln(4)}{1 \cdot \ln(3)} = \frac{\ln(4)}{\ln(3)} \]

### Step 5: Use the change of base formula
Using the change of base formula for logarithms, we can rewrite:
\[ \frac{\ln(4)}{\ln(3)} = \log_3(4) \]

### Final Answer
Thus, the limit is:
\[ \lim_{n \to \infty} \frac{4^{1/n} - 1}{3^{1/n} - 1} = \log_3(4) \]

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