The value of `lim_(x to oo)((1+3x)/(2+3x))^((1-sqrt(x))/(1-x))` is

Options

A. 0
B. `-1`
C. `e`
D. `1`

Answer

To solve the limit \( \lim_{x \to \infty} \left( \frac{1 + 3x}{2 + 3x} \right)^{\frac{1 - \sqrt{x}}{1 - x}} \), we will break it down into two parts: the base and the exponent.

### Step 1: Evaluate the base limit
We start with the base of the expression:
\[ f_1 = \lim_{x \to \infty} \frac{1 + 3x}{2 + 3x} \]

To simplify this, we can factor out \(3x\) from both the numerator and the denominator:
\[ f_1 = \lim_{x \to \infty} \frac{3x(1 + \frac{1}{3x})}{3x(1 + \frac{2}{3x})} = \lim_{x \to \infty} \frac{1 + \frac{1}{3x}}{1 + \frac{2}{3x}} \]

As \(x\) approaches infinity, \(\frac{1}{3x}\) and \(\frac{2}{3x}\) both approach 0:
\[ f_1 = \frac{1 + 0}{1 + 0} = 1 \]

### Step 2: Evaluate the exponent limit
Next, we evaluate the exponent:
\[ f_2 = \lim_{x \to \infty} \frac{1 - \sqrt{x}}{1 - x} \]

To simplify this expression, we can rewrite it:
\[ f_2 = \lim_{x \to \infty} \frac{1 - \sqrt{x}}{1 - x} = \lim_{x \to \infty} \frac{-(\sqrt{x} - 1)}{-(x - 1)} = \lim_{x \to \infty} \frac{\sqrt{x} - 1}{x - 1} \]

Now, we can factor out \(\sqrt{x}\) from the numerator and \(x\) from the denominator:
\[ f_2 = \lim_{x \to \infty} \frac{\sqrt{x}(1 - \frac{1}{\sqrt{x}})}{x(1 - \frac{1}{x})} = \lim_{x \to \infty} \frac{1 - \frac{1}{\sqrt{x}}}{\sqrt{x}(1 - \frac{1}{x})} \]

As \(x\) approaches infinity, \(\frac{1}{\sqrt{x}}\) and \(\frac{1}{x}\) both approach 0:
\[ f_2 = \lim_{x \to \infty} \frac{1 - 0}{\sqrt{x}(1 - 0)} = \lim_{x \to \infty} \frac{1}{\sqrt{x}} = 0 \]

### Step 3: Combine the results
Now we can combine the results of \(f_1\) and \(f_2\):
\[ \lim_{x \to \infty} \left( \frac{1 + 3x}{2 + 3x} \right)^{\frac{1 - \sqrt{x}}{1 - x}} = 1^0 \]

Since any non-zero number raised to the power of 0 is equal to 1, we conclude:
\[ \lim_{x \to \infty} \left( \frac{1 + 3x}{2 + 3x} \right)^{\frac{1 - \sqrt{x}}{1 - x}} = 1 \]

### Final Answer:
The value of the limit is \( \boxed{1} \).