If `f(x) = sqrt((x + sinx)/(x + cos^(2)x))`. Then `lim_(x to infty) f(x)` is:

Options

A. 0
B. `infty`
C. 1
D. none of these

Answer

To find the limit of the function \( f(x) = \sqrt{\frac{x + \sin x}{x + \cos^2 x}} \) as \( x \) approaches infinity, we can follow these steps:

### Step 1: Analyze the function
We start with the function:
\[ f(x) = \sqrt{\frac{x + \sin x}{x + \cos^2 x}} \]

### Step 2: Simplify the expression
As \( x \) approaches infinity, both \( \sin x \) and \( \cos^2 x \) oscillate between fixed values (-1 to 1 for \( \sin x \) and 0 to 1 for \( \cos^2 x \)). Therefore, their contributions become negligible compared to \( x \).

We can rewrite the function:
\[ f(x) = \sqrt{\frac{x(1 + \frac{\sin x}{x})}{x(1 + \frac{\cos^2 x}{x})}} = \sqrt{\frac{1 + \frac{\sin x}{x}}{1 + \frac{\cos^2 x}{x}}} \]

### Step 3: Evaluate the limits of the terms
Now we need to evaluate the limits of the terms \( \frac{\sin x}{x} \) and \( \frac{\cos^2 x}{x} \) as \( x \) approaches infinity.

1. **Limit of \( \frac{\sin x}{x} \)**:
\[ \lim_{x \to \infty} \frac{\sin x}{x} = 0 \]
(since \( \sin x \) is bounded between -1 and 1)

2. **Limit of \( \frac{\cos^2 x}{x} \)**:
\[ \lim_{x \to \infty} \frac{\cos^2 x}{x} = 0 \]
(for the same reason, \( \cos^2 x \) is also bounded between 0 and 1)

### Step 4: Substitute the limits back into the function
Now we can substitute these limits back into our simplified function:
\[ \lim_{x \to \infty} f(x) = \sqrt{\frac{1 + 0}{1 + 0}} = \sqrt{1} = 1 \]

### Conclusion
Thus, the limit of \( f(x) \) as \( x \) approaches infinity is:
\[ \lim_{x \to \infty} f(x) = 1 \]

### Final Answer
The limit is \( 1 \).

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