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

Options

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

Answer

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

### Step 1: Simplify the expression inside the limit
We start with the expression:
\[ \frac{x}{3+x} \]
As \( x \) approaches infinity, we can simplify this expression:
\[ \frac{x}{3+x} = \frac{x}{x(1 + \frac{3}{x})} = \frac{1}{1 + \frac{3}{x}} \]
As \( x \to \infty \), \( \frac{3}{x} \to 0 \), so:
\[ \frac{x}{3+x} \to \frac{1}{1+0} = 1 \]

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

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

### Step 4: Apply the limit definition
Using the limit definition \( \lim_{m \to \infty} \left( 1 + \frac{\lambda}{m} \right)^{m} = e^{\lambda} \), we can rewrite:
\[ \lim_{x \to \infty} \left( 1 - \frac{3}{x+3} \right)^{(x+3)} = e^{-3} \]

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

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