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The problem in the image asks for the value of the following limit as \( x \to \infty \):

\[
\lim_{x \to \infty} \left( \sqrt{x^2 + 2x + 1} - \sqrt{x^2 - 3x + 3} \right)
\]

### Solution:

1. **Factor the expressions inside the square roots**: As \( x \to \infty \), we should aim to simplify both square root expressions by factoring out \( x^2 \), the dominant term.

\[
\sqrt{x^2 + 2x + 1} = x \sqrt{1 + \frac{2}{x} + \frac{1}{x^2}}
\]
\[
\sqrt{x^2 - 3x + 3} = x \sqrt{1 - \frac{3}{x} + \frac{3}{x^2}}
\]

2. **Expand the square roots using a binomial approximation**: Since \( \frac{1}{x} \) and \( \frac{1}{x^2} \) become very small as \( x \to \infty \), we can use the approximation \( \sqrt{1 + u} \approx 1 + \frac{u}{2} \) for small \( u \).

- For \( \sqrt{1 + \frac{2}{x} + \frac{1}{x^2}} \):
\[
\sqrt{1 + \frac{2}{x} + \frac{1}{x^2}} \approx 1 + \frac{1}{x}
\]

- For \( \sqrt{1 - \frac{3}{x} + \frac{3}{x^2}} \):
\[
\sqrt{1 - \frac{3}{x} + \frac{3}{x^2}} \approx 1 - \frac{3}{2x}
\]

3. **Substitute the approximations back into the expression**:

\[
\sqrt{x^2 + 2x + 1} \approx x \left( 1 + \frac{1}{x} \right) = x + 1
\]
\[
\sqrt{x^2 - 3x + 3} \approx x \left( 1 - \frac{3}{2x} \right) = x - \frac{3}{2}
\]

4. **Simplify the difference**:

\[
\sqrt{x^2 + 2x + 1} - \sqrt{x^2 - 3x + 3} \approx (x + 1) - \left( x - \frac{3}{2} \right)
\]
\[
= x + 1 - x + \frac{3}{2} = 1 + \frac{3}{2} = \frac{5}{2}
\]

Thus, the value of the limit is:

\[
\boxed{\frac{5}{2}}
\]

Would you like further details or have any questions?

### Related questions:
1. How does the binomial approximation work for square roots?
2. What other techniques can be used for finding limits involving square roots?
3. How do we handle limits as \( x \to 0 \) for similar expressions?
4. What if the expressions inside the square roots had higher powers of \( x \)?
5. How does L'Hopital's Rule apply in this case?

### Tip:
When solving limits involving square roots, factor out the dominant term and use approximations to simplify the expression as \( x \to \infty \).

Nilai dari \lim_{x \to \infty} \left( \sqrt{x^2 + 2x + 1} - \sqrt{x^2 - 3x + 3} \right)

This problem demonstrates how to evaluate the limit of a difference of two square root functions as x approaches infinity. It uses binomial approximation for square roots and simplifies the expression step-by-step. Suitable for students in calculus courses.

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