The given mathematical expression is:

$\lim_{x \to 0} \frac{e^{\arctan(x)/x} - e}{x}$

To evaluate this limit, let us proceed step by step:

Step 1: Examine the exponential term

As $x \to 0$, the term $\frac{\arctan(x)}{x}$ behaves as follows:

• For small values of $x$, we know that $\arctan(x) \sim x$ (since the Maclaurin series for $\arctan(x)$ starts as $\arctan(x) = x - \frac{x^3}{3} + \mathcal{O}(x^5)$).

Thus:

$\frac{\arctan(x)}{x} \to 1 \quad \text{as} \quad x \to 0.$

So, the exponential term $e^{\frac{\arctan(x)}{x}} \to e$.

Step 2: Write the numerator as a difference

To simplify the term $e^{\frac{\arctan(x)}{x}} - e$, we can factor out $e$ since both terms approach $e$ as $x \to 0$. Rewrite:

$e^{\frac{\arctan(x)}{x}} - e = e \left( e^{\frac{\arctan(x)}{x} - 1} - 1 \right).$

Step 3: Approximate $e^u$ near $u = 0$

For small $u$, the exponential function can be approximated as:

$e^u \approx 1 + u \quad \text{(first-order Taylor series)}.$

Here $u = \frac{\arctan(x)}{x} - 1$. Since $\frac{\arctan(x)}{x} \to 1$, let us define:

$u = \frac{\arctan(x)}{x} - 1.$

Therefore:

$e^{\frac{\arctan(x)}{x} - 1} - 1 \approx \frac{\arctan(x)}{x} - 1.$

Step 4: Analyze $\frac{\arctan(x)}{x} - 1$ near $x = 0$

From the Taylor series of $\arctan(x)$, we have:

$\arctan(x) = x - \frac{x^3}{3} + \mathcal{O}(x^5).$

Thus:

$\frac{\arctan(x)}{x} = 1 - \frac{x^2}{3} + \mathcal{O}(x^4).$

Subtracting 1:

$\frac{\arctan(x)}{x} - 1 = -\frac{x^2}{3} + \mathcal{O}(x^4).$

Step 5: Substitute into the numerator

Now substitute $\frac{\arctan(x)}{x} - 1$ into the numerator:

$e^{\frac{\arctan(x)}{x}} - e \approx e \left( -\frac{x^2}{3} \right).$

Step 6: Divide by $x$ in the denominator

The original limit is:

$\lim_{x \to 0} \frac{e^{\frac{\arctan(x)}{x}} - e}{x}.$

Substitute the simplified numerator:

$\frac{e^{\frac{\arctan(x)}{x}} - e}{x} \approx \frac{e \left( -\frac{x^2}{3} \right)}{x}.$

Simplify:

$\frac{e \left( -\frac{x^2}{3} \right)}{x} = -\frac{e}{3} x.$

Step 7: Take the limit as $x \to 0$

As $x \to 0$, the term $-\frac{e}{3} x \to 0$. Therefore:

$\lim_{x \to 0} \frac{e^{\frac{\arctan(x)}{x}} - e}{x} = 0.$

Final Answer:

$\boxed{0}$

Do you want me to elaborate further on any steps or clarify any part of the solution? Let me know!

Related Questions:

1. How do you find the limit of a similar exponential function involving $x \to 0$?
2. What is the Taylor series of $\arctan(x)$, and how is it used to approximate small values?
3. How can we use L'Hôpital's Rule to evaluate similar indeterminate forms?
4. Why does the exponential approximation $e^u \approx 1 + u$ hold for small $u$?
5. What happens to the limit if $x \to \infty$ instead of $x \to 0$?

Tip: When handling limits involving exponential functions, Taylor series expansions often simplify indeterminate forms. Always check the behavior near $x = 0$ carefully!