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The function \( f(x) \) is defined as a piecewise function:

\[
f(x) = 
\begin{cases} 
\text{tan}\sqrt{1-x}, & \text{for } x \leq 1 \\
\frac{\sin^2 \sqrt{x^2 - 1}}{3x^2 - 3}, & \text{for } x > 1
\end{cases}
\]

You are asked to find:
1. \( f(1) \)
2. \( \lim_{x \to 1^+} f(x) \)
3. \( \lim_{x \to 1^-} f(x) \)

### 1. Finding \( f(1) \):
Since \( f(x) = \text{tan}\sqrt{1-x} \) for \( x \leq 1 \), we use this expression to evaluate \( f(1) \):
\[
f(1) = \text{tan}\sqrt{1 - 1} = \text{tan}(0) = 0
\]

Thus, \( f(1) = 0 \).

### 2. Finding \( \lim_{x \to 1^+} f(x) \):
For \( x > 1 \), the function is:
\[
f(x) = \frac{\sin^2 \sqrt{x^2 - 1}}{3x^2 - 3}
\]

We need to compute the limit as \( x \to 1^+ \):
As \( x \to 1^+ \), we observe that:
\[
\sqrt{x^2 - 1} \to \sqrt{1^2 - 1} = \sqrt{0} = 0
\]
and
\[
3x^2 - 3 = 3(1^2) - 3 = 3 - 3 = 0
\]
Thus, we are dealing with a \( \frac{0}{0} \) indeterminate form. To resolve this, we can apply L'Hopital's Rule by differentiating the numerator and denominator with respect to \( x \).

#### Applying L'Hopital's Rule:
The numerator is \( \sin^2(\sqrt{x^2 - 1}) \), and the denominator is \( 3x^2 - 3 \). Differentiating both:

- Derivative of the numerator \( \sin^2(\sqrt{x^2 - 1}) \):
\[
\frac{d}{dx} \sin^2(\sqrt{x^2 - 1}) = 2 \sin(\sqrt{x^2 - 1}) \cdot \cos(\sqrt{x^2 - 1}) \cdot \frac{d}{dx}(\sqrt{x^2 - 1}) = 2 \sin(\sqrt{x^2 - 1}) \cos(\sqrt{x^2 - 1}) \cdot \frac{x}{\sqrt{x^2 - 1}}
\]
At \( x = 1^+ \), this expression approaches 0 because \( \sin(\sqrt{x^2 - 1}) \) approaches 0.

- Derivative of the denominator \( 3x^2 - 3 \):
\[
\frac{d}{dx}(3x^2 - 3) = 6x
\]
At \( x = 1^+ \), this equals 6.

Thus, applying L'Hopital's Rule:
\[
\lim_{x \to 1^+} f(x) = \frac{0}{6} = 0
\]

### 3. Finding \( \lim_{x \to 1^-} f(x) \):
For \( x \leq 1 \), the function is \( f(x) = \text{tan}(\sqrt{1 - x}) \).

As \( x \to 1^- \):
\[
\sqrt{1 - x} \to \sqrt{0} = 0
\]
and
\[
\text{tan}(0) = 0
\]

Thus, \( \lim_{x \to 1^-} f(x) = 0 \).

### Conclusion:
- \( f(1) = 0 \)
- \( \lim_{x \to 1^+} f(x) = 0 \)
- \( \lim_{x \to 1^-} f(x) = 0 \)

Since the left-hand limit, right-hand limit, and the value of the function at \( x = 1 \) are all equal, the function is continuous at \( x = 1 \).

---

Would you like to explore more details or ask any further questions? Here are five related questions you might find interesting:
1. How is L'Hopital's Rule applied to other indeterminate forms like \( \frac{\infty}{\infty} \)?
2. Can we use different methods, such as series expansion, to solve the limit problem?
3. How does the continuity of a piecewise function get affected by different kinds of discontinuities?
4. What happens if the limits on both sides of \( x = 1 \) were different?
5. How do we determine higher-order differentiability at points where a function is piecewise?

**Tip**: Always check the differentiability of functions when using L'Hopital's Rule, especially at critical points like \( x = 1 \).

Tính f(1); lim x → 1+ f(x); lim x → 1- f(x) for the piecewise function f(x) defined as tan(√(1 - x)) for x ≤ 1, and sin²(√(x² - 1)) / (3x² - 3) for x > 1.

Solve the limit problem for a piecewise function defined as tan(√(1 - x)) for x ≤ 1 and sin²(√(x² - 1)) / (3x² - 3) for x > 1. The problem involves L'Hopital's Rule and continuity. Suitable for Grades 11-12 or early college students.

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