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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 $0$.

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.