You have the expression \(\frac{x \csc(x) + 1}{x \csc(x)}\) and you are interested in evaluating its limit as \(x\) approaches 0.

Let's break it down step-by-step:

### Expression:
\[
f(x) = \frac{x \csc(x) + 1}{x \csc(x)}
\]

The function involves the cosecant function, \(\csc(x) = \frac{1}{\sin(x)}\). So, we can rewrite the expression using this identity:

\[
f(x) = \frac{x \left(\frac{1}{\sin(x)}\right) + 1}{x \left(\frac{1}{\sin(x)}\right)} = \frac{\frac{x}{\sin(x)} + 1}{\frac{x}{\sin(x)}}
\]

### Limit as \(x \to 0\):
We now consider the limit of this expression as \(x\) approaches 0.

\[
\lim_{x \to 0} \frac{\frac{x}{\sin(x)} + 1}{\frac{x}{\sin(x)}}
\]

It is known that \(\frac{x}{\sin(x)} \to 1\) as \(x \to 0\). Therefore, as \(x\) approaches 0, the expression simplifies:

\[
\lim_{x \to 0} \frac{1 + 1}{1} = \frac{2}{1} = 2
\]

### Final Answer:
The limit of \(\frac{x \csc(x) + 1}{x \csc(x)}\) as \(x\) approaches 0 is **2**.

Would you like further details or have any other questions? Here are 5 related questions for further exploration:

1. What is the limit of \(\frac{x}{\sin(x)}\) as \(x\) approaches 0?
2. How does L'Hôpital's Rule apply to limits involving trigonometric functions?
3. What is the significance of small-angle approximations for trigonometric functions?
4. How can you compute the limit of more complex trigonometric functions at 0?
5. What is the limit of \(\frac{\sin(x)}{x}\) as \(x\) approaches 0?

**Tip:** When working with limits involving trigonometric functions, small-angle approximations like \(\sin(x) \approx x\) for small \(x\) are extremely helpful!

Learn how to compute the limit of (xcscx + 1) / xcscx as x approaches 0. This calculus problem involves understanding the behavior of trigonometric functions at small angles, applying limit theorems, and using key formulas like csc(x) = 1/sin(x). Suitable for Grades 11-12 and early college students.

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