We need to find the limit of the function $\frac{e^x + 2x - 1}{2e^x + 1}$ as $x$ approaches infinity.

1. Problem Description

We need to find the limit of the function $\frac{e^x + 2x - 1}{2e^x + 1}$ as $x$ approaches infinity.

2. Solution Steps

We are given the limit:

$$$$

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

$$$$

As $x$ approaches infinity, $e^x$ also approaches infinity. Both the numerator and denominator approach infinity, so we have an indeterminate form of type $\frac{\infty}{\infty}$. We can apply L'Hopital's Rule. L'Hopital's Rule states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the limit on the right exists.

First, we find the derivatives of the numerator and the denominator:

$$$$

\frac{d}{dx}(e^x + 2x - 1) = e^x + 2

$$$$

$$$$

\frac{d}{dx}(2e^x + 1) = 2e^x

$$$$

So, applying L'Hopital's Rule once, we get:

$$$$

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

$$$$

This is still of the form $\frac{\infty}{\infty}$, so we can apply L'Hopital's Rule again.

$$$$

\frac{d}{dx}(e^x + 2) = e^x

$$$$

$$$$

\frac{d}{dx}(2e^x) = 2e^x

$$$$

So, applying L'Hopital's Rule a second time, we get:

$$$$

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

$$$$

We can simplify the fraction:

$$$$

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

$$$$

Alternatively, we can divide the numerator and denominator by $e^x$ from the beginning:

$$$$

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

$$$$

As $x \to \infty$, $\frac{1}{e^x} \to 0$. Also, $\lim_{x \to \infty} \frac{2x}{e^x} = 0$ by L'Hopital's Rule (since $\lim_{x \to \infty} \frac{2}{e^x} = 0$).

$$$$

\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

$\frac{1}{2}$