`f(x)=("ln"(x^(2)+e^(x)))/("ln"(x^(4)+e^(2x)))`. Then `lim_(x to oo)` f(x) is equal to

Options

A. `(2a)/(pi)`
B. `-(2a)/(pi)`
C. `(4a)/(pi)`
D. `-(4a)/(pi)`

Explanation

To find the limit of the function \( f(x) = \frac{\ln(x^2 + e^x)}{\ln(x^4 + e^{2x})} \) as \( x \) approaches infinity, we can follow these steps:

### Step 1: Identify the form of the limit
As \( x \to \infty \):
- The term \( e^x \) grows much faster than \( x^2 \), so \( x^2 + e^x \) approaches \( e^x \).
- Similarly, \( e^{2x} \) grows much faster than \( x^4 \), so \( x^4 + e^{2x} \) approaches \( e^{2x} \).

Thus, we can rewrite the limit:
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{\ln(e^x)}{\ln(e^{2x})} \]

### Step 2: Simplify the logarithms
Using the property of logarithms \( \ln(e^a) = a \):
\[ \ln(e^x) = x \quad \text{and} \quad \ln(e^{2x}) = 2x \]

### Step 3: Substitute back into the limit
Now we substitute these simplifications back into the limit:
\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x}{2x} \]

### Step 4: Simplify the expression
This simplifies to:
\[ \lim_{x \to \infty} \frac{x}{2x} = \lim_{x \to \infty} \frac{1}{2} = \frac{1}{2} \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to \infty} f(x) = \frac{1}{2} \]