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))`
The value of `lim_(xtooo)f(x)` is

Options

A. 0
B. 2
C. 5
D. 1

Answer

To solve the problem, we will evaluate the limit step by step.

Given:
- \( g(x) = \frac{1}{x} \)
- \( h(x) = x^2 + 2x + (\lambda + 1) \)
- \( u(x) = \frac{1}{x} + \cos\left(\frac{1}{x^2}\right) \)

We need to find:
\[ f(x) = \lim_{n \to \infty} \frac{x^{2n+1} g(x) + h(x)}{x^{2n} + 3x u(x)} \]

### Step 1: Substitute \( g(x) \) and \( h(x) \) into \( f(x) \)

Substituting \( g(x) \) and \( h(x) \):
\[ f(x) = \lim_{n \to \infty} \frac{x^{2n+1} \cdot \frac{1}{x} + (x^2 + 2x + (\lambda + 1))}{x^{2n} + 3x \left(\frac{1}{x} + \cos\left(\frac{1}{x^2}\right)\right)} \]

### Step 2: Simplify the expression

This simplifies to:
\[ f(x) = \lim_{n \to \infty} \frac{x^{2n} + x^2 + 2x + (\lambda + 1)}{x^{2n} + 3 + 3x \cos\left(\frac{1}{x^2}\right)} \]

### Step 3: Divide numerator and denominator by \( x^{2n} \)

Now, divide both the numerator and the denominator by \( x^{2n} \):
\[ f(x) = \lim_{n \to \infty} \frac{1 + \frac{x^2}{x^{2n}} + \frac{2x}{x^{2n}} + \frac{\lambda + 1}{x^{2n}}}{1 + \frac{3}{x^{2n}} + \frac{3 \cos\left(\frac{1}{x^2}\right)}{x^{2n-1}}} \]

### Step 4: Evaluate the limit as \( n \to \infty \)

As \( n \to \infty \), the terms \( \frac{x^2}{x^{2n}}, \frac{2x}{x^{2n}}, \frac{\lambda + 1}{x^{2n}}, \frac{3}{x^{2n}}, \) and \( \frac{3 \cos\left(\frac{1}{x^2}\right)}{x^{2n-1}} \) all approach 0. Thus, we have:
\[ f(x) = \frac{1 + 0 + 0 + 0}{1 + 0 + 0} = 1 \]

### Step 5: Final limit

Therefore, we conclude:
\[ \lim_{x \to \infty} f(x) = 1 \]

### Final Answer

The value of \( \lim_{x \to \infty} f(x) \) is \( \boxed{1} \).