To evaluate the limit of $\frac{\sin(x)}{\cos^2(x)}$ as $x$ approaches 1, we proceed as follows:

$\lim_{x \to 1} \frac{\sin(x)}{\cos^2(x)}$

Since $\sin(x)$ and $\cos(x)$ are continuous functions, we can directly substitute $x = 1$ into the expression:

$\sin(1) \approx 0.8415 \quad \text{and} \quad \cos(1) \approx 0.5403$

Therefore:

$\cos^2(1) = (\cos(1))^2 \approx (0.5403)^2 \approx 0.2919$

Now we can compute the limit:

$\lim_{x \to 1} \frac{\sin(x)}{\cos^2(x)} \approx \frac{0.8415}{0.2919} \approx 2.882$

So,

$\lim_{x \to 1} \frac{\sin(x)}{\cos^2(x)} \approx 2.882$

Do you want more details or have any questions?

Here are 5 more questions you might want to ask:

1. What is the limit of $\frac{\sin(x)}{x}$ as $x$ approaches 0?
2. How do you find the derivative of $\tan(x)$?
3. What is the value of $\lim_{x \to 0} \frac{1 - \cos(x)}{x^2}$?
4. Can you explain the concept of continuity for trigonometric functions?
5. What is the integral of $\tan(x)$ with respect to $x$?

Tip: For limits involving trigonometric functions, remember that direct substitution is often a helpful first step if the functions are continuous at the point of interest.