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 $x \to \infty$.