We are asked to evaluate the limit: $\lim_{x\to\infty} \frac{e^x + 2x - 1}{2e^x + 1}$

1. Problem Description

We are asked to evaluate the limit:

$\lim_{x\to\infty} \frac{e^x + 2x - 1}{2e^x + 1}$

2. Solution Steps

To evaluate the limit, we can divide both the numerator and the denominator by $e^x$.

$\lim_{x\to\infty} \frac{e^x + 2x - 1}{2e^x + 1} = \lim_{x\to\infty} \frac{\frac{e^x}{e^x} + \frac{2x}{e^x} - \frac{1}{e^x}}{\frac{2e^x}{e^x} + \frac{1}{e^x}} = \lim_{x\to\infty} \frac{1 + \frac{2x}{e^x} - \frac{1}{e^x}}{2 + \frac{1}{e^x}}$

Now we analyze the limits of the terms $\frac{2x}{e^x}$ and $\frac{1}{e^x}$ as $x \to \infty$.

$\lim_{x\to\infty} \frac{1}{e^x} = 0$

To evaluate $\lim_{x\to\infty} \frac{2x}{e^x}$, we can use L'Hopital's rule, since it is in the indeterminate form $\frac{\infty}{\infty}$.

Taking the derivative of the numerator and denominator, we get:

$\lim_{x\to\infty} \frac{2x}{e^x} = \lim_{x\to\infty} \frac{2}{e^x} = 0$

Thus we have:

$\lim_{x\to\infty} \frac{1 + \frac{2x}{e^x} - \frac{1}{e^x}}{2 + \frac{1}{e^x}} = \frac{1 + 0 - 0}{2 + 0} = \frac{1}{2}$

3. Final Answer

The final answer is $\frac{1}{2}$