`"lt"_(x to infty) (3^(x+1) -5^(x+1))/(3^(x) - 5^(x))=`

Options

A. `-5`
B. `1//5`
C. 5
D. none

Answer

To solve the limit

\[ \lim_{x \to \infty} \frac{3^{x+1} - 5^{x+1}}{3^x - 5^x} \]

we will follow these steps:

### Step 1: Rewrite the expression
We can rewrite the limit as:

\[ \lim_{x \to \infty} \frac{3 \cdot 3^x - 5 \cdot 5^x}{3^x - 5^x} \]

This allows us to factor out terms more easily.

### Step 2: Factor out the dominant term
In both the numerator and the denominator, the term \(5^x\) will dominate as \(x\) approaches infinity. Therefore, we can factor \(5^x\) out of both the numerator and the denominator:

\[ = \lim_{x \to \infty} \frac{5^x \left( \frac{3 \cdot 3^x}{5^x} - 5 \right)}{5^x \left( \frac{3^x}{5^x} - 1 \right)} \]

### Step 3: Simplify the expression
Now we can cancel \(5^x\) from the numerator and denominator:

\[ = \lim_{x \to \infty} \frac{\frac{3 \cdot 3^x}{5^x} - 5}{\frac{3^x}{5^x} - 1} \]

### Step 4: Substitute \( \left( \frac{3}{5} \right)^x \)
Let \(y = \left( \frac{3}{5} \right)^x\). As \(x \to \infty\), \(y \to 0\):

\[ = \lim_{y \to 0} \frac{3y - 5}{y - 1} \]

### Step 5: Evaluate the limit
Now substituting \(y = 0\):

\[ = \frac{3(0) - 5}{0 - 1} = \frac{-5}{-1} = 5 \]

### Final Answer
Thus, the limit is:

\[ \boxed{5} \]