The value of `lim_(xrarroo)((x+6)/(x+1))^(x+4)`, is

Options

A. e
B. `e^2`
C. `e^4`
D. ` e^5`

Answer

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

### Step 1: Rewrite the expression
We start by rewriting the expression inside the limit:
\[ \frac{x+6}{x+1} = \frac{x(1 + \frac{6}{x})}{x(1 + \frac{1}{x})} = \frac{1 + \frac{6}{x}}{1 + \frac{1}{x}} \]
Thus, we can express our limit as:
\[ \lim_{x \to \infty} \left( \frac{1 + \frac{6}{x}}{1 + \frac{1}{x}} \right)^{x+4} \]

### Step 2: Evaluate the limit of the base
As \( x \to \infty \), both \( \frac{6}{x} \) and \( \frac{1}{x} \) approach 0. Therefore:
\[ \frac{1 + \frac{6}{x}}{1 + \frac{1}{x}} \to \frac{1 + 0}{1 + 0} = 1 \]
This means we are dealing with an indeterminate form \( 1^{\infty} \).

### Step 3: Apply the exponential limit property
To resolve the indeterminate form \( 1^{\infty} \), we can use the property:
\[ a^b = e^{b \ln a} \]
Thus, we rewrite our limit as:
\[ \lim_{x \to \infty} \left( \frac{1 + \frac{6}{x}}{1 + \frac{1}{x}} \right)^{x+4} = e^{\lim_{x \to \infty} (x+4) \ln\left(\frac{1 + \frac{6}{x}}{1 + \frac{1}{x}}\right)} \]

### Step 4: Simplify the logarithm
Next, we need to simplify the logarithm:
\[ \ln\left(\frac{1 + \frac{6}{x}}{1 + \frac{1}{x}}\right) = \ln(1 + \frac{6}{x}) - \ln(1 + \frac{1}{x}) \]
Using the approximation \( \ln(1 + u) \approx u \) for small \( u \):
\[ \ln(1 + \frac{6}{x}) \approx \frac{6}{x}, \quad \ln(1 + \frac{1}{x}) \approx \frac{1}{x} \]
Thus:
\[ \ln\left(\frac{1 + \frac{6}{x}}{1 + \frac{1}{x}}\right) \approx \frac{6}{x} - \frac{1}{x} = \frac{5}{x} \]

### Step 5: Substitute back into the limit
Now we substitute this back into our limit:
\[ \lim_{x \to \infty} (x+4) \ln\left(\frac{1 + \frac{6}{x}}{1 + \frac{1}{x}}\right) \approx \lim_{x \to \infty} (x+4) \cdot \frac{5}{x} = \lim_{x \to \infty} \left(5 + \frac{20}{x}\right) = 5 \]

### Step 6: Final result
Thus, we have:
\[ \lim_{x \to \infty} \left( \frac{x+6}{x+1} \right)^{x+4} = e^5 \]

### Conclusion
The value of the limit is:
\[ \boxed{e^5} \]