We are given the limit:

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

The goal is to find values of $\delta$ that correspond to a given $\epsilon$.

Definition of Limit:

For the limit $\lim_{{x \to 0}} f(x) = L$, we say that the limit equals $L$ if, for every $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $0 < |x - 0| < \delta$, the following holds:

$|f(x) - L| < \epsilon$

In this case, $f(x) = \frac{e^x - 1}{x}$ and $L = 1$.

We need to satisfy:

$\left| \frac{e^x - 1}{x} - 1 \right| < \epsilon$

Simplifying the Expression:

$\left| \frac{e^x - 1}{x} - 1 \right| = \left| \frac{e^x - 1 - x}{x} \right|$ So we need:

$\left| \frac{e^x - 1 - x}{x} \right| < \epsilon$

Now, we must find $\delta$ in terms of $\epsilon$.

Step-by-Step Solution:

1. We need to solve for $x$ such that $\left| \frac{e^x - 1 - x}{x} \right| < \epsilon$.

2. Approximate $e^x$ using the series expansion for small $x$:

$e^x = 1 + x + \frac{x^2}{2} + O(x^3)$

Thus, $e^x - 1 - x \approx \frac{x^2}{2}$, and we have:

$\left| \frac{\frac{x^2}{2}}{x} \right| = \left| \frac{x}{2} \right|$

3. We now need $\left| \frac{x}{2} \right| < \epsilon$, which gives:

$|x| < 2\epsilon$

4. Therefore, we can choose $\delta = 2\epsilon$.

For $\epsilon = 0.5$:

$\delta = 2 \times 0.5 = 1.0$

Thus, the corresponding $\delta$ for $\epsilon = 0.5$ is $\delta = 1.0$.

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Would you like a more detailed explanation or have any other questions?

Here are 5 related questions to explore:

1. How do you find limits using the epsilon-delta definition?
2. What is the Taylor series expansion of $e^x$?
3. How do you solve for $\delta$ in terms of $\epsilon$ in general?
4. What are common limits involving exponential functions?
5. How does the approximation $e^x \approx 1 + x$ for small $x$ help in solving limits?

Tip: When using the epsilon-delta definition, approximating complex functions with Taylor series can greatly simplify the problem.