Evaluate the following limits :
`Lim_(x to oo) ((x+6)/(x+1))^(x+4)`

Answer

To evaluate the limit \( \lim_{x \to \infty} \left( \frac{x+6}{x+1} \right)^{x+4} \), we can follow these steps:

### Step 1: Set Up the Limit
Let \( L = \lim_{x \to \infty} \left( \frac{x+6}{x+1} \right)^{x+4} \).

### Step 2: Take the Natural Logarithm
Taking the natural logarithm of both sides, we have:
\[ \log L = \lim_{x \to \infty} (x+4) \log\left( \frac{x+6}{x+1} \right) \]

### Step 3: Simplify the Logarithm
We can simplify the logarithm:
\[ \log\left( \frac{x+6}{x+1} \right) = \log(x+6) - \log(x+1) \]
As \( x \to \infty \), both \( \log(x+6) \) and \( \log(x+1) \) approach \( \log x \). Thus,
\[ \log\left( \frac{x+6}{x+1} \right) \sim \log x - \log x = \log\left( 1 + \frac{5}{x+1} \right) \]

### Step 4: Use the Logarithmic Approximation
Using the approximation \( \log(1 + u) \approx u \) for small \( u \):
\[ \log\left( 1 + \frac{5}{x+1} \right) \approx \frac{5}{x+1} \]
So,
\[ \log\left( \frac{x+6}{x+1} \right) \approx \frac{5}{x+1} \]

### Step 5: Substitute Back into the Limit
Now substituting back into our limit:
\[ \log L = \lim_{x \to \infty} (x+4) \cdot \frac{5}{x+1} \]

### Step 6: Simplify the Expression
This simplifies to:
\[ \log L = \lim_{x \to \infty} \frac{5(x+4)}{x+1} = \lim_{x \to \infty} \frac{5x + 20}{x + 1} \]
Dividing numerator and denominator by \( x \):
\[ \log L = \lim_{x \to \infty} \frac{5 + \frac{20}{x}}{1 + \frac{1}{x}} = \frac{5 + 0}{1 + 0} = 5 \]

### Step 7: Exponentiate to Find L
Thus, we have:
\[ \log L = 5 \implies L = e^5 \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to \infty} \left( \frac{x+6}{x+1} \right)^{x+4} = e^5 \]

---