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

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.

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

Learn how to evaluate the limit $$ \lim_{\overrightarrow{x}0}(\tan x - x)^{-1} $$ using Taylor series expansion. This calculus problem explores the behavior of tan(x) around zero and involves key concepts like limits and series approximations. Suitable for undergraduate calculus students.

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