The problem asks to solve the limit:

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

and determine how positive $x$ needs to be for:

$\frac{{e^x}}{{2 + 2e^x}}$

to be within one millionth of the limiting value $L$.

Step 1: Solve the Limit

We start by simplifying the expression as $x \to \infty$:

$\frac{e^x}{2 + 2e^x}$

When $x \to \infty$, the exponential term $e^x$ grows very large, while the constant $2$ remains unchanged. To make the limit more obvious, we factor $e^x$ out of the denominator:

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

Dividing the numerator and denominator by $e^x$:

$\frac{e^x}{2(1 + e^x)} = \frac{1}{2 \left( \frac{1}{e^x} + 1 \right)}$

As $x \to \infty$, $\frac{1}{e^x} \to 0$, so the expression simplifies to:

$\frac{1}{2(0 + 1)} = \frac{1}{2}$

Thus, the limiting value $L$ is:

$L = \frac{1}{2}$

Step 2: Determine $x$ for Precision

Now, we need to find how large $x$ must be so that:

$\left| \frac{e^x}{2 + 2e^x} - \frac{1}{2} \right| < 10^{-6}$

Simplify this expression:

$\frac{e^x}{2 + 2e^x} - \frac{1}{2} = \frac{1}{2} \left( \frac{1 - \frac{1}{e^x}}{1 + \frac{1}{e^x}} \right)$

We need:

$\left| \frac{1 - \frac{1}{e^x}}{1 + \frac{1}{e^x}} \right| < 2 \times 10^{-6}$

For large $x$, $\frac{1}{e^x}$ becomes small. Setting $\frac{1}{e^x} = \epsilon$, we approximate:

$\left| \frac{1 - \epsilon}{1 + \epsilon} \right| \approx 1 - 2\epsilon$

So we want:

$2\epsilon < 2 \times 10^{-6}$

This simplifies to:

$\epsilon < 10^{-6}$

Thus:

$\frac{1}{e^x} < 10^{-6} \quad \Rightarrow \quad e^x > 10^6$

Taking the natural logarithm of both sides:

$x > \ln(10^6) = 6 \ln(10) \approx 13.815$

Therefore, $x$ needs to be greater than approximately $13.815$ to ensure the expression is within one millionth of $\frac{1}{2}$.

Summary:

• The limit is $\frac{1}{2}$.
• $x > 13.815$ ensures the expression is within one millionth of the limit.

Would you like further clarification or details on any step?

Related Questions:

1. What is the general method for solving limits at infinity?
2. How does the exponential function $e^x$ behave as $x \to \infty$?
3. What is the significance of dividing by the largest term in the denominator when calculating limits?
4. How can logarithms be used to solve inequalities involving exponential functions?
5. What other functions grow similarly to $e^x$ as $x \to \infty$?

Tip:

When finding limits involving exponential functions at infinity, it helps to simplify by factoring out the dominant exponential term from both the numerator and denominator.