We are asked to find the limit of the expression $\sec(x) \cot(x)$ as $x$ approaches 0:

$\lim_{x \to 0} \sec(x) \cot(x)$

Step 1: Rewrite $\sec(x)$ and $\cot(x)$ in terms of basic trigonometric functions

We know that:

• $\sec(x) = \frac{1}{\cos(x)}$
• $\cot(x) = \frac{\cos(x)}{\sin(x)}$

Substituting these into the expression, we get:

$\sec(x) \cot(x) = \frac{1}{\cos(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{1}{\sin(x)}$

Step 2: Simplify the expression

Thus, the original expression reduces to:

$\frac{1}{\sin(x)}$

Step 3: Take the limit as $x \to 0$

We now need to find the limit:

$\lim_{x \to 0} \frac{1}{\sin(x)}$

Since $\sin(x) \to 0$ as $x \to 0$, the limit of $\frac{1}{\sin(x)}$ approaches $\infty$. Therefore:

$\lim_{x \to 0} \frac{1}{\sin(x)} = \infty$

Conclusion:

The limit of $\sec(x) \cot(x)$ as $x$ approaches 0 is $\infty$.

Would you like further details on this solution or have any questions?

Here are 5 related questions to consider:

1. What is the limit of $\frac{1}{\tan(x)}$ as $x \to 0$?
2. How does the behavior of trigonometric functions around 0 affect limits?
3. What is the significance of limits approaching infinity in real-world scenarios?
4. Can L'Hopital's Rule be applied to solve similar limits, and how?
5. What are some common limits involving $\sec(x)$ and $\cot(x)$?

Tip: When dealing with trigonometric limits, rewriting the functions in simpler terms (like sine and cosine) often makes the limit easier to compute.