What is `lim_(xtooo) ((x)/(3+x))^(3x)` equal to?

Options

A. e
B. `e^(3)`
C. `e^(-9)`
D. `e^(9)`

Explanation

To solve the limit \( \lim_{x \to \infty} \left( \frac{x}{3+x} \right)^{3x} \), we will follow these steps:

### Step 1: Simplify the expression inside the limit
We start with the expression:
\[ \frac{x}{3+x} \]
As \( x \to \infty \), we can divide the numerator and denominator by \( x \):
\[ \frac{x}{3+x} = \frac{x/x}{3/x + x/x} = \frac{1}{\frac{3}{x} + 1} \]
As \( x \to \infty \), \( \frac{3}{x} \to 0 \), so:
\[ \frac{x}{3+x} \to \frac{1}{0 + 1} = 1 \]

### Step 2: Rewrite the limit
Now we rewrite the limit:
\[ \lim_{x \to \infty} \left( \frac{x}{3+x} \right)^{3x} = \lim_{x \to \infty} \left( 1 - \frac{3}{x+3} \right)^{3x} \]
This is because \( \frac{x}{3+x} = 1 - \frac{3}{x+3} \).

### Step 3: Use the exponential limit property
We can express the limit in the form suitable for applying the exponential limit:
\[ \lim_{x \to \infty} \left( 1 - \frac{3}{x+3} \right)^{3x} = \lim_{x \to \infty} \left( \left( 1 - \frac{3}{x+3} \right)^{(x+3)} \right)^{\frac{3x}{x+3}} \]
As \( x \to \infty \), \( \frac{3x}{x+3} \to 3 \).

### Step 4: Evaluate the inner limit
Now we focus on the inner limit:
\[ \lim_{x \to \infty} \left( 1 - \frac{3}{x+3} \right)^{(x+3)} \]
This limit approaches \( e^{-3} \) as \( x \to \infty \) because it is of the form \( \left( 1 + \frac{a}{n} \right)^n \) where \( a = -3 \).

### Step 5: Combine the results
Thus, we have:
\[ \lim_{x \to \infty} \left( 1 - \frac{3}{x+3} \right)^{(x+3)} = e^{-3} \]
And since \( \frac{3x}{x+3} \to 3 \):
\[ \lim_{x \to \infty} \left( 1 - \frac{3}{x+3} \right)^{3x} = \left( e^{-3} \right)^3 = e^{-9} \]

### Final Result
Therefore, the limit is:
\[ \lim_{x \to \infty} \left( \frac{x}{3+x} \right)^{3x} = e^{-9} \]