`lim_(xrarroo) (x^(3)+3x^(2)+6x+5)/(x^(3)+x+2)`

Answer

To find the limit as \( x \) approaches infinity for the expression

\[ \lim_{x \to \infty} \frac{x^3 + 3x^2 + 6x + 5}{x^3 + x + 2} \]

we can follow these steps:

### Step 1: Identify the highest power of \( x \)
In both the numerator and the denominator, the highest power of \( x \) is \( x^3 \).

### Step 2: Divide each term by \( x^3 \)
To simplify the expression, divide every term in the numerator and the denominator by \( x^3 \):

\[ \lim_{x \to \infty} \frac{\frac{x^3}{x^3} + \frac{3x^2}{x^3} + \frac{6x}{x^3} + \frac{5}{x^3}}{\frac{x^3}{x^3} + \frac{x}{x^3} + \frac{2}{x^3}} \]

This simplifies to:

\[ \lim_{x \to \infty} \frac{1 + \frac{3}{x} + \frac{6}{x^2} + \frac{5}{x^3}}{1 + \frac{1}{x^2} + \frac{2}{x^3}} \]

### Step 3: Evaluate the limit as \( x \) approaches infinity
As \( x \) approaches infinity, the terms \( \frac{3}{x} \), \( \frac{6}{x^2} \), \( \frac{5}{x^3} \), \( \frac{1}{x^2} \), and \( \frac{2}{x^3} \) all approach \( 0 \). Therefore, we can simplify the limit to:

\[ \lim_{x \to \infty} \frac{1 + 0 + 0 + 0}{1 + 0 + 0} = \frac{1}{1} \]

### Step 4: Conclusion
Thus, the limit is:

\[ \lim_{x \to \infty} \frac{x^3 + 3x^2 + 6x + 5}{x^3 + x + 2} = 1 \]