Evaluate `lim_(x to sqrt(3)) (3x^(8) + x^(7) - 11x^(6) - 2x^(5) - 9x^(4) - x^(3) + 35x^(2) + 6x + 30)/(x^(5) - 2x^(4) + 4x^(2) - 9x + 6)`

Explanation

To evaluate the limit

\[ \lim_{x \to \sqrt{3}} \frac{3x^8 + x^7 - 11x^6 - 2x^5 - 9x^4 - x^3 + 35x^2 + 6x + 30}{x^5 - 2x^4 + 4x^2 - 9x + 6}, \]

we will follow these steps:

### Step 1: Substitute \(x = \sqrt{3}\)

First, we substitute \(x = \sqrt{3}\) into the numerator and denominator to check if we get an indeterminate form \( \frac{0}{0} \).

**Numerator:**
\[ 3(\sqrt{3})^8 + (\sqrt{3})^7 - 11(\sqrt{3})^6 - 2(\sqrt{3})^5 - 9(\sqrt{3})^4 - (\sqrt{3})^3 + 35(\sqrt{3})^2 + 6(\sqrt{3}) + 30 \]
Calculating each term:
- \(3(\sqrt{3})^8 = 3 \cdot 81 = 243\)
- \((\sqrt{3})^7 = 3^{3.5} = 3 \cdot 9 = 27\sqrt{3}\)
- \(-11(\sqrt{3})^6 = -11 \cdot 27 = -297\)
- \(-2(\sqrt{3})^5 = -2 \cdot 9\sqrt{3} = -18\sqrt{3}\)
- \(-9(\sqrt{3})^4 = -9 \cdot 9 = -81\)
- \(-(\sqrt{3})^3 = -3\sqrt{3}\)
- \(35(\sqrt{3})^2 = 35 \cdot 3 = 105\)
- \(6(\sqrt{3}) = 6\sqrt{3}\)
- \(30\)

Combining these gives us:
\[ 243 + 27\sqrt{3} - 297 - 18\sqrt{3} - 81 - 3\sqrt{3} + 105 + 6\sqrt{3} + 30 \]
This simplifies to:
\[ (243 - 297 - 81 + 105 + 30) + (27\sqrt{3} - 18\sqrt{3} - 3\sqrt{3} + 6\sqrt{3}) = 0 + 0 = 0 \]

**Denominator:**
\[ (\sqrt{3})^5 - 2(\sqrt{3})^4 + 4(\sqrt{3})^2 - 9(\sqrt{3}) + 6 \]
Calculating each term:
- \((\sqrt{3})^5 = 9\sqrt{3}\)
- \(-2(\sqrt{3})^4 = -2 \cdot 9 = -18\)
- \(4(\sqrt{3})^2 = 4 \cdot 3 = 12\)
- \(-9(\sqrt{3})\)
- \(6\)

Combining these gives us:
\[ 9\sqrt{3} - 18 + 12 - 9\sqrt{3} + 6 = 0 \]

Since both the numerator and denominator equal zero, we have an indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]

if \( \lim_{x \to a} f(x) = 0 \) and \( \lim_{x \to a} g(x) = 0 \).

### Step 3: Differentiate the Numerator and Denominator

**Numerator:**
\[ f(x) = 3x^8 + x^7 - 11x^6 - 2x^5 - 9x^4 - x^3 + 35x^2 + 6x + 30 \]
Differentiating:
\[ f'(x) = 24x^7 + 7x^6 - 66x^5 - 10x^4 - 36x^3 - 3x^2 + 70x + 6 \]

**Denominator:**
\[ g(x) = x^5 - 2x^4 + 4x^2 - 9x + 6 \]
Differentiating:
\[ g'(x) = 5x^4 - 8x^3 + 8x - 9 \]

### Step 4: Evaluate the Limit Again

Now we evaluate the limit again:

\[ \lim_{x \to \sqrt{3}} \frac{f'(x)}{g'(x)} \]

Substituting \(x = \sqrt{3}\):

**Numerator:**
\[ f'(\sqrt{3}) = 24(\sqrt{3})^7 + 7(\sqrt{3})^6 - 66(\sqrt{3})^5 - 10(\sqrt{3})^4 - 36(\sqrt{3})^3 + 70(\sqrt{3}) + 6 \]

Calculating each term:
- \(24(\sqrt{3})^7 = 24 \cdot 27\sqrt{3} = 648\sqrt{3}\)
- \(7(\sqrt{3})^6 = 7 \cdot 27 = 189\)
- \(-66(\sqrt{3})^5 = -66 \cdot 9\sqrt{3} = -594\sqrt{3}\)
- \(-10(\sqrt{3})^4 = -10 \cdot 9 = -90\)
- \(-36(\sqrt{3})^3 = -36 \cdot 3\sqrt{3} = -108\sqrt{3}\)
- \(70(\sqrt{3}) = 70\sqrt{3}\)
- \(6\)

Combining these gives us:
\[ (189 - 90 + 6) + (648\sqrt{3} - 594\sqrt{3} - 108\sqrt{3} + 70\sqrt{3}) = 105 + 0 = 105 \]

**Denominator:**
\[ g'(\sqrt{3}) = 5(\sqrt{3})^4 - 8(\sqrt{3})^3 + 8(\sqrt{3}) - 9 \]

Calculating each term:
- \(5(\sqrt{3})^4 = 5 \cdot 9 = 45\)
- \(-8(\sqrt{3})^3 = -8 \cdot 3\sqrt{3} = -24\sqrt{3}\)
- \(8(\sqrt{3}) = 8\sqrt{3}\)
- \(-9\)

Combining these gives us:
\[ 45 - 9 + (8\sqrt{3} - 24\sqrt{3}) = 36 - 16\sqrt{3} \]

### Step 5: Final Limit Calculation

Now we have:

\[ \lim_{x \to \sqrt{3}} \frac{f'(\sqrt{3})}{g'(\sqrt{3})} = \frac{105}{36 - 16\sqrt{3}} \]

This is our final answer.