The expression given is:

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

This limit is evaluated as $x$ approaches 0 from the left side, which is indicated by $x \to 0^-$. Let's analyze the behavior of the expression as $x$ approaches 0 from the left (i.e., $x < 0$).

Step 1: Analyze $e^{1/x}$ as $x \to 0^-$

When $x$ is a small negative number, $\frac{1}{x}$ becomes a large negative number. Therefore, $e^{1/x}$ approaches 0 because the exponential function $e^y$ approaches 0 as $y$ becomes very negative.

Step 2: Substitute the behavior into the limit

As $x \to 0^-$, $e^{1/x}$ tends to 0. Thus, the expression simplifies as:

$\frac{e^{1/x} - 1}{e^{1/x} + 1} \approx \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1$

Conclusion

Therefore, the limit is:

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

Would you like to explore any related concepts or have any other questions? Here are a few related topics you might find interesting:

1. What happens to the limit if $x$ approaches 0 from the right (i.e., $x \to 0^+$)?
2. How does the behavior of the exponential function change as its exponent becomes very large or very small?
3. Can this type of limit be evaluated using L'Hôpital's Rule? Why or why not?
4. What are the key differences between limits approaching from the left and from the right?
5. How do exponential functions behave compared to polynomial or logarithmic functions as their input approaches infinity or negative infinity?

Tip: When evaluating limits, especially as $x$ approaches a particular value, always consider the direction from which $x$ is approaching, as this can significantly impact the result.