`lim_(x to oo) (( 1+x+x^(3)))/((ln x)^(3))` is equal to

Options

A. 2
B. `e^(2)`
C. `e^(-2)`
D. Not defined

Answer

To solve the limit \( \lim_{x \to \infty} \frac{1 + x + x^3}{(\ln x)^3} \), we will follow these steps:

### Step 1: Analyze the limit
As \( x \) approaches infinity, both the numerator and denominator approach infinity, leading to the indeterminate form \( \frac{\infty}{\infty} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:

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

We will differentiate the numerator and the denominator.

### Step 3: Differentiate the numerator and denominator
- **Numerator**: \( f(x) = 1 + x + x^3 \)
- \( f'(x) = 0 + 1 + 3x^2 = 1 + 3x^2 \)

- **Denominator**: \( g(x) = (\ln x)^3 \)
- Using the chain rule:
- \( g'(x) = 3(\ln x)^2 \cdot \frac{1}{x} = \frac{3(\ln x)^2}{x} \)

### Step 4: Rewrite the limit
Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to \infty} \frac{1 + 3x^2}{\frac{3(\ln x)^2}{x}} = \lim_{x \to \infty} \frac{(1 + 3x^2) \cdot x}{3(\ln x)^2} \]

### Step 5: Simplify the expression
This simplifies to:

\[ \lim_{x \to \infty} \frac{x + 3x^3}{3(\ln x)^2} \]

### Step 6: Analyze the new limit
As \( x \to \infty \), the term \( 3x^3 \) dominates the numerator, so we can focus on that:

\[ \lim_{x \to \infty} \frac{3x^3}{3(\ln x)^2} = \lim_{x \to \infty} \frac{x^3}{(\ln x)^2} \]

### Step 7: Apply L'Hôpital's Rule again
This is still an indeterminate form \( \frac{\infty}{\infty} \), so we apply L'Hôpital's Rule again:

- **Numerator**: \( f(x) = x^3 \)
- \( f'(x) = 3x^2 \)

- **Denominator**: \( g(x) = (\ln x)^2 \)
- \( g'(x) = 2(\ln x) \cdot \frac{1}{x} = \frac{2\ln x}{x} \)

Now we rewrite the limit:

\[ \lim_{x \to \infty} \frac{3x^2}{\frac{2\ln x}{x}} = \lim_{x \to \infty} \frac{3x^3}{2\ln x} \]

### Step 8: Apply L'Hôpital's Rule again
This is still \( \frac{\infty}{\infty} \), so we apply L'Hôpital's Rule again:

- **Numerator**: \( f(x) = 3x^3 \)
- \( f'(x) = 9x^2 \)

- **Denominator**: \( g(x) = 2\ln x \)
- \( g'(x) = \frac{2}{x} \)

Now we rewrite the limit:

\[ \lim_{x \to \infty} \frac{9x^2}{\frac{2}{x}} = \lim_{x \to \infty} \frac{9x^3}{2} \]

### Step 9: Evaluate the limit
As \( x \to \infty \), \( \frac{9x^3}{2} \) approaches infinity.

### Final Result
Thus, we conclude that:

\[ \lim_{x \to \infty} \frac{1 + x + x^3}{(\ln x)^3} = \infty \]