`lim_(xrarr oo)((x^2+4x-3)/(x^2-2x+5))^x` is equal to

Options

A. `e^6`
B. `e^-6`
C. `e^2`
D. `e^4`

Explanation

To solve the limit problem \( \lim_{x \to \infty} \left( \frac{x^2 + 4x - 3}{x^2 - 2x + 5} \right)^x \), we can follow these steps:

### Step 1: Simplify the expression inside the limit
We start by simplifying the fraction inside the limit:

\[ \frac{x^2 + 4x - 3}{x^2 - 2x + 5} \]

As \( x \) approaches infinity, we can factor out \( x^2 \) from both the numerator and the denominator:

\[ \frac{x^2(1 + \frac{4}{x} - \frac{3}{x^2})}{x^2(1 - \frac{2}{x} + \frac{5}{x^2})} \]

This simplifies to:

\[ \frac{1 + \frac{4}{x} - \frac{3}{x^2}}{1 - \frac{2}{x} + \frac{5}{x^2}} \]

### Step 2: Evaluate the limit of the simplified expression
Now, we take the limit of this expression as \( x \) approaches infinity:

\[ \lim_{x \to \infty} \frac{1 + \frac{4}{x} - \frac{3}{x^2}}{1 - \frac{2}{x} + \frac{5}{x^2}} = \frac{1 + 0 - 0}{1 - 0 + 0} = 1 \]

### Step 3: Rewrite the limit in exponential form
Since we have \( \left( \frac{x^2 + 4x - 3}{x^2 - 2x + 5} \right)^x \) approaching \( 1^x \), which is an indeterminate form \( 1^\infty \), we can rewrite it using the exponential limit:

\[ \lim_{x \to \infty} \left( \frac{x^2 + 4x - 3}{x^2 - 2x + 5} \right)^x = e^{\lim_{x \to \infty} \left( \frac{x^2 + 4x - 3}{x^2 - 2x + 5} - 1 \right) x} \]

### Step 4: Find the limit of the exponent
Next, we need to find the limit of the exponent:

\[ \lim_{x \to \infty} \left( \frac{x^2 + 4x - 3 - (x^2 - 2x + 5)}{x^2 - 2x + 5} \right) x \]

This simplifies to:

\[ \lim_{x \to \infty} \left( \frac{6x - 8}{x^2 - 2x + 5} \right) x \]

### Step 5: Simplify the limit further
Now, we can simplify this limit:

\[ = \lim_{x \to \infty} \frac{6x^2 - 8x}{x^2 - 2x + 5} \]

Dividing the numerator and the denominator by \( x^2 \):

\[ = \lim_{x \to \infty} \frac{6 - \frac{8}{x}}{1 - \frac{2}{x} + \frac{5}{x^2}} = \frac{6 - 0}{1 - 0 + 0} = 6 \]

### Step 6: Final result
Thus, we have:

\[ \lim_{x \to \infty} \left( \frac{x^2 + 4x - 3}{x^2 - 2x + 5} \right)^x = e^6 \]

So the final answer is:

\[ \boxed{e^6} \]