Given `g(x)=(1/x), h(x)=x^(2)+2x+(lamda+1)` and `u(x)=1/x+cos(1/(x^(2)))`
Let `f(x)=lim_(ntooo)(x^(2n+1)g(x)+h(x))/(x^(2n)+3x.u(x))`
If `lim+(xto2)f(x)=I`, then [I] (where [.] denotes the greatest integer function), is equal to :

Options

A. 0
B. 1
C. -1
D. 2

Answer

To solve the problem step by step, we start with the given functions and the limit we need to evaluate.

### Step 1: Write down the functions
We have:
- \( g(x) = \frac{1}{x} \)
- \( h(x) = x^2 + 2x + (\lambda + 1) \)
- \( u(x) = \frac{1}{x} + \cos\left(\frac{1}{x^2}\right) \)

### Step 2: Write the expression for \( f(x) \)
The function \( f(x) \) is defined as:
\[ f(x) = \lim_{n \to \infty} \frac{x^{2n+1} g(x) + h(x)}{x^{2n} + 3x u(x)} \]

### Step 3: Substitute \( g(x) \) and \( u(x) \) into \( f(x) \)
Substituting \( g(x) \) and \( u(x) \):
\[ f(x) = \lim_{n \to \infty} \frac{x^{2n+1} \cdot \frac{1}{x} + h(x)}{x^{2n} + 3x \left(\frac{1}{x} + \cos\left(\frac{1}{x^2}\right)\right)} \]
This simplifies to:
\[ f(x) = \lim_{n \to \infty} \frac{x^{2n} + h(x)}{x^{2n} + 3 + 3x \cos\left(\frac{1}{x^2}\right)} \]

### Step 4: Simplify \( h(x) \)
Substituting \( h(x) \):
\[ h(x) = x^2 + 2x + (\lambda + 1) \]
So,
\[ f(x) = \lim_{n \to \infty} \frac{x^{2n} + x^2 + 2x + (\lambda + 1)}{x^{2n} + 3 + 3x \cos\left(\frac{1}{x^2}\right)} \]

### Step 5: Analyze the limit as \( n \to \infty \)
As \( n \to \infty \), the terms \( x^{2n} \) dominate both the numerator and the denominator:
\[ f(x) = \lim_{n \to \infty} \frac{1 + \frac{x^2 + 2x + (\lambda + 1)}{x^{2n}}}{1 + \frac{3 + 3x \cos\left(\frac{1}{x^2}\right)}{x^{2n}}} \]
As \( n \to \infty \), the fractions involving \( x^{2n} \) approach 0:
\[ f(x) = \frac{1 + 0}{1 + 0} = 1 \]

### Step 6: Evaluate \( \lim_{x \to 2} f(x) \)
Since \( f(x) = 1 \) for all \( x \) in the limit as \( n \to \infty \):
\[ \lim_{x \to 2} f(x) = 1 \]

### Step 7: Find the greatest integer function value
Let \( I = \lim_{x \to 2} f(x) = 1 \). Therefore, the greatest integer function \( [I] \) is:
\[ [I] = [1] = 1 \]

### Final Answer
The value of \( [I] \) is \( 1 \).