The given expression is:

$\lim_{\overrightarrow{x}0}\left(\tan x\,-x\right)^{-1}$

Step-by-Step Solution:

1. Analyze the Limit Expression: We are tasked with evaluating the behavior of $\left( \tan x - x \right)^{-1}$ as $x \to 0$.

2. Taylor Series Expansion: Around $x = 0$, we can approximate $\tan x$ using its Taylor series expansion: $\tan x = x + \frac{x^3}{3} + O(x^5)$ Therefore, for small $x$, $\tan x - x = \left(x + \frac{x^3}{3} + O(x^5)\right) - x = \frac{x^3}{3} + O(x^5)$

3. Substitute the Expansion: Substituting this approximation into the original expression: $\left( \tan x - x \right)^{-1} \approx \left( \frac{x^3}{3} \right)^{-1} = \frac{3}{x^3}$

4. Evaluate the Limit: Now, evaluate the limit as $x \to 0$: $\lim_{x \to 0} \frac{3}{x^3}$ As $x \to 0$, $\frac{3}{x^3} \to \infty$.

Thus, the limit does not exist (the expression tends to infinity).

$\boxed{\infty}$

Would you like more details or have any questions?

Here are 5 related questions to expand on this:

1. How do we derive the Taylor series of $\tan x$?
2. What other functions exhibit similar behavior near $x = 0$?
3. How would the limit change if the function were $\left(\sin x - x\right)^{-1}$?
4. What happens if we take the limit $\lim_{x \to 0} \frac{\tan x - x}{x^3}$?
5. Can we evaluate this limit using L'Hopital's Rule instead of series expansion?

Tip: Taylor series are a powerful tool for approximating functions near specific points, especially for evaluating tricky limits.