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!