evaluate the limit $$ \lim_{x\rarr\infty}\frac{e^{\frac{1}{x}}-1}{\frac{1}{x}} $$

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.

Evaluate the limit $$ \lim_{x \to \infty} \frac{e^{\frac{1}{x}} - 1}{\frac{1}{x}} $$

Learn to evaluate the limit $ \lim_{x \to \infty} \frac{e^{\frac{1}{x}} - 1}{\frac{1}{x}} $ using L'Hôpital's Rule. This problem covers essential calculus concepts like exponential limits, substitution techniques, and L'Hôpital's Rule, making it ideal for advanced high school or college calculus students.

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