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### Problem in the Image:

The problem is:

\[ \lim_{x \to 0} \left(\frac{1}{x} - \frac{1}{\sin x}\right) = \lim_{x \to 0} \left(\frac{\sin x - x}{x \sin x}\right) \]

We need to evaluate the limit:

\[
\lim_{x \to 0} \left(\frac{\sin x - x}{x \sin x}\right)
\]

### Solution:

To solve this limit, we can use the Taylor series expansion for \(\sin x\) around \(x = 0\):

\[
\sin x = x - \frac{x^3}{6} + \mathcal{O}(x^5)
\]

So, 

\[
\sin x - x = \left(x - \frac{x^3}{6} + \mathcal{O}(x^5)\right) - x = -\frac{x^3}{6} + \mathcal{O}(x^5)
\]

Substituting this into the limit:

\[
\lim_{x \to 0} \left(\frac{\sin x - x}{x \sin x}\right) = \lim_{x \to 0} \frac{-\frac{x^3}{6} + \mathcal{O}(x^5)}{x \left(x - \frac{x^3}{6} + \mathcal{O}(x^5)\right)}
\]

Simplifying the denominator:

\[
x \sin x = x \left(x - \frac{x^3}{6} + \mathcal{O}(x^5)\right) = x^2 - \frac{x^4}{6} + \mathcal{O}(x^6)
\]

Now the limit becomes:

\[
\lim_{x \to 0} \frac{-\frac{x^3}{6}}{x^2 - \frac{x^4}{6}} 
\]

Further simplification gives:

\[
\lim_{x \to 0} \frac{-\frac{x^3}{6}}{x^2 \left(1 - \frac{x^2}{6}\right)} = \lim_{x \to 0} \frac{-\frac{x^3}{6}}{x^2} \cdot \frac{1}{1 - \frac{x^2}{6}} = \lim_{x \to 0} \frac{-\frac{x}{6}}{1 - \frac{x^2}{6}}
\]

As \(x \to 0\), the limit simplifies to:

\[
\lim_{x \to 0} -\frac{x}{6} = 0
\]

### Conclusion:

The limit evaluates to \(0\).

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Do you have any questions or would you like more details on any part of the solution? Here are a few related questions you might consider:

1. How does the Taylor series help in evaluating limits involving trigonometric functions?
2. What are the common limits involving \(\sin(x)\) and how are they derived?
3. How can L'Hôpital's rule be used to evaluate this limit?
4. Why does the higher-order term \(\mathcal{O}(x^5)\) become negligible as \(x \to 0\)?
5. What other methods could be used to solve this limit problem?

**Tip:** When dealing with trigonometric limits, Taylor series expansions can be very useful for simplification and understanding the behavior near a specific point.

Evaluate the limit \( \lim_{x \to 0} \left(\frac{1}{x} - \frac{1}{\sin x}\right) = \lim_{x \to 0} \left(\frac{\sin x - x}{x \sin x}\right) \)

Learn how to evaluate the limit \(\lim_{x \to 0} \left(\frac{\sin x - x}{x \sin x}\right)\) using the Taylor series expansion for \(\sin x\). This calculus problem covers essential concepts like limits and Taylor series, suitable for students in Grades 11-12.

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