To evaluate the limit

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

we can start by rewriting it in a more convenient form to apply L'Hôpital's Rule.

Step 1: Rewrite the limit

As $x \to \infty$, $\frac{1}{x} \to 0$. Thus, we can rewrite the limit in terms of a new variable $u = \frac{1}{x}$. Then as $x \to \infty$, $u \to 0^+$. Substituting $u = \frac{1}{x}$ into the limit, we get:

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

Step 2: Apply L'Hôpital's Rule

The limit now has the form $\frac{0}{0}$ as $u \to 0$, so we can apply L'Hôpital's Rule by differentiating the numerator and denominator with respect to $u$:

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

Step 3: Evaluate the limit

Now, substituting $u = 0$ into the remaining expression:

$\lim_{u \to 0} e^u = e^0 = 1.$

Final Answer

Therefore, the limit is

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

Would you like further details on this solution?

Here are 5 related questions you might find helpful:

1. How would you evaluate $\lim_{x \to 0} \frac{e^{2x} - 1}{x}$?
2. What is $\lim_{x \to \infty} \frac{e^{\frac{2}{x}} - 1}{\frac{1}{x}}$?
3. How do you solve $\lim_{x \to 0} \frac{\ln(1 + x)}{x}$?
4. How would you approach $\lim_{x \to 0} \frac{\sin x}{x}$ using similar techniques?
5. How does the expansion of $e^u$ help in finding limits like this?

Tip: Using substitutions like $u = \frac{1}{x}$ can simplify limits, making it easier to apply L'Hôpital's Rule.