Let `f(x)=(log_e(x^2+e^x))/(log_e(x^4+e^2x))`. If `lim_(xrarr oo) f(x)=l` and `lim_(xrarr-oo)f(x)=m`, then

Options

A. `l=m`
B. `l=2m`
C. `2l=m`
D. `l+m=0`

Explanation

To solve the problem, we need to find the limits of the function \( f(x) = \frac{\log_e(x^2 + e^x)}{\log_e(x^4 + e^{2x})} \) as \( x \) approaches infinity and negative infinity.

### Step 1: Evaluate the limit as \( x \to \infty \)

We start with:
\[ f(x) = \frac{\log_e(x^2 + e^x)}{\log_e(x^4 + e^{2x})} \]

As \( x \to \infty \), both the numerator and denominator approach infinity:
- In the numerator, \( e^x \) dominates \( x^2 \), so \( \log_e(x^2 + e^x) \approx \log_e(e^x) = x \).
- In the denominator, \( e^{2x} \) dominates \( x^4 \), so \( \log_e(x^4 + e^{2x}) \approx \log_e(e^{2x}) = 2x \).

Thus, we can write:
\[ f(x) \approx \frac{x}{2x} = \frac{1}{2} \]

So,
\[ \lim_{x \to \infty} f(x) = \frac{1}{2} \]

### Step 2: Evaluate the limit as \( x \to -\infty \)

Now we consider the limit as \( x \to -\infty \):
\[ f(x) = \frac{\log_e(x^2 + e^x)}{\log_e(x^4 + e^{2x})} \]

As \( x \to -\infty \):
- In the numerator, \( e^x \) approaches 0, so \( x^2 \) dominates, and \( \log_e(x^2 + e^x) \approx \log_e(x^2) = 2\log_e(-x) \).
- In the denominator, \( e^{2x} \) also approaches 0, so \( x^4 \) dominates, and \( \log_e(x^4 + e^{2x}) \approx \log_e(x^4) = 4\log_e(-x) \).

Thus, we can write:
\[ f(x) \approx \frac{2\log_e(-x)}{4\log_e(-x)} = \frac{1}{2} \]

So,
\[ \lim_{x \to -\infty} f(x) = \frac{1}{2} \]

### Conclusion

We have found:
\[ l = \lim_{x \to \infty} f(x) = \frac{1}{2} \]
\[ m = \lim_{x \to -\infty} f(x) = \frac{1}{2} \]

Thus, \( l = m \).