The value of `lim_(xtooo) ((20^(x)-1)/(19(5^(x))))^(1//x)` is ________.

Options

A. 3
B. 1
C. `(2)/(3)`
D. `(3)/(2)`

Explanation

To solve the limit \( \lim_{x \to \infty} \left( \frac{20^x - 1}{19 \cdot 5^x} \right)^{\frac{1}{x}} \), we will follow these steps:

### Step 1: Set up the limit
Let \( L = \lim_{x \to \infty} \left( \frac{20^x - 1}{19 \cdot 5^x} \right)^{\frac{1}{x}} \).

### Step 2: Take the natural logarithm
Taking the natural logarithm of both sides gives:
\[ \ln L = \lim_{x \to \infty} \frac{1}{x} \ln \left( \frac{20^x - 1}{19 \cdot 5^x} \right) \]

### Step 3: Simplify the logarithm
Using the properties of logarithms, we can rewrite it as:
\[ \ln L = \lim_{x \to \infty} \frac{1}{x} \left( \ln(20^x - 1) - \ln(19) - \ln(5^x) \right) \]

### Step 4: Further simplify the expression
We can express \( \ln(5^x) \) as \( x \ln(5) \):
\[ \ln L = \lim_{x \to \infty} \frac{1}{x} \left( \ln(20^x - 1) - \ln(19) - x \ln(5) \right) \]

### Step 5: Analyze \( \ln(20^x - 1) \)
As \( x \to \infty \), \( 20^x - 1 \) approaches \( 20^x \), so:
\[ \ln(20^x - 1) \sim \ln(20^x) = x \ln(20) \]
Thus, we can replace \( \ln(20^x - 1) \) with \( x \ln(20) \) in the limit:
\[ \ln L = \lim_{x \to \infty} \frac{1}{x} \left( x \ln(20) - \ln(19) - x \ln(5) \right) \]

### Step 6: Factor out \( x \)
This simplifies to:
\[ \ln L = \lim_{x \to \infty} \left( \ln(20) - \frac{\ln(19)}{x} - \ln(5) \right) \]

### Step 7: Evaluate the limit
As \( x \to \infty \), the term \( \frac{\ln(19)}{x} \) approaches \( 0 \):
\[ \ln L = \ln(20) - \ln(5) = \ln\left(\frac{20}{5}\right) = \ln(4) \]

### Step 8: Exponentiate to find \( L \)
Exponentiating both sides gives:
\[ L = e^{\ln(4)} = 4 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{4} \]