If `x gt0` and g is bounded function then `lim_(ntooo)(f(x)e^(nx)+g(x))/(e^(nx)+1)`

Options

A. `f(x)`
B. `g(x)`
C. `0`
D. None of these

Answer

To solve the limit problem, we need to evaluate the limit:

\[ \lim_{n \to \infty} \frac{f(x)e^{nx} + g(x)}{e^{nx} + 1} \]

where \( x > 0 \) and \( g(x) \) is a bounded function.

### Step-by-step Solution:

1. **Identify the Form of the Limit**:
As \( n \to \infty \), both \( e^{nx} \) terms in the numerator and denominator grow very large. Therefore, we can see that we have an indeterminate form of type \( \frac{\infty}{\infty} \).

2. **Factor Out \( e^{nx} \)**:
We can simplify the expression by factoring \( e^{nx} \) out of both the numerator and the denominator:

\[ = \lim_{n \to \infty} \frac{e^{nx}(f(x) + \frac{g(x)}{e^{nx}})}{e^{nx}(1 + \frac{1}{e^{nx}})} \]

3. **Cancel \( e^{nx} \)**:
Since \( e^{nx} \) is common in both the numerator and denominator, we can cancel it (as long as \( e^{nx} \neq 0 \), which it isn't for \( x > 0 \)):

\[ = \lim_{n \to \infty} \frac{f(x) + \frac{g(x)}{e^{nx}}}{1 + \frac{1}{e^{nx}}} \]

4. **Evaluate the Limit**:
As \( n \to \infty \), \( \frac{g(x)}{e^{nx}} \) approaches 0 because \( g(x) \) is bounded (let's say \( |g(x)| \leq M \) for some constant \( M \)), and \( e^{nx} \) grows without bound. Similarly, \( \frac{1}{e^{nx}} \) also approaches 0. Thus, we have:

\[ = \frac{f(x) + 0}{1 + 0} = f(x) \]

5. **Conclusion**:
Therefore, the limit is:

\[ \lim_{n \to \infty} \frac{f(x)e^{nx} + g(x)}{e^{nx} + 1} = f(x) \]

### Final Answer:
The limit is \( f(x) \).