$\lim _{x \rightarrow 0} \frac{\mathrm{e}^{\tan x}-\mathrm{e}^x}{\tan x-x}=$

Mean Value Theorem Application

The limit evaluates as $\lim_{x \to 0} \frac{e^{\tan x} - e^x}{\tan x - x}$. Consider $f(t) = e^t$, which is continuous and differentiable on $\mathbb{R}$, with $f'(t) = e^t$.

For $x$ near $0$, apply the Mean Value Theorem on the interval between $x$ and $\tan x$. There exists $c$ in $(x, \tan x)$ such that $\frac{e^{\tan x} - e^x}{\tan x - x} = e^c$.

As $x \to 0$, both $x$ and $\tan x$ approach $0$, so $c \to 0$. Therefore, the limit becomes $\lim_{c \to 0} e^c = e^0 = 1$.

Taylor Series Verification

Using expansions near $x=0$: $e^u = 1 + u + \frac{u^2}{2} + \frac{u^3}{6} + O(u^4)$ and $\tan x = x + \frac{x^3}{3} + O(x^5)$.

Compute $e^{\tan x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{2} + O(x^4)$ and $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)$. Subtract: $e^{\tan x} - e^x = \frac{x^3}{3} + O(x^4)$.

Denominator is $\tan x - x = \frac{x^3}{3} + O(x^5)$. Form ratio: $\frac{\frac{x^3}{3} + O(x^4)}{\frac{x^3}{3} + O(x^5)} = \frac{\frac{1}{3} + O(x)}{\frac{1}{3} + O(x^2)} \to 1$ as $x \to 0$.

Both methods yield $\boxed{1}$.