To evaluate the limit $\lim_{x \to \infty} \left( x - \sqrt{x^2 + 1} \right)$, we can use the technique of rationalizing the expression. This involves multiplying and dividing by the conjugate of the expression inside the limit.

Here's the step-by-step process:

1. Start with the given limit: $\lim_{x \to \infty} \left( x - \sqrt{x^2 + 1} \right)$

2. Multiply and divide by the conjugate of the expression: $\lim_{x \to \infty} \left( x - \sqrt{x^2 + 1} \right) \cdot \frac{x + \sqrt{x^2 + 1}}{x + \sqrt{x^2 + 1}}$

3. Simplify the numerator using the difference of squares: $= \lim_{x \to \infty} \frac{(x - \sqrt{x^2 + 1})(x + \sqrt{x^2 + 1})}{x + \sqrt{x^2 + 1}}$ $= \lim_{x \to \infty} \frac{x^2 - (x^2 + 1)}{x + \sqrt{x^2 + 1}}$ $= \lim_{x \to \infty} \frac{x^2 - x^2 - 1}{x + \sqrt{x^2 + 1}}$ $= \lim_{x \to \infty} \frac{-1}{x + \sqrt{x^2 + 1}}$

4. Simplify the expression further: $= \lim_{x \to \infty} \frac{-1}{x + \sqrt{x^2 + 1}}$

As $x \to \infty$, both $x$ and $\sqrt{x^2 + 1}$ grow large. Specifically, $\sqrt{x^2 + 1}$ behaves like $x$ because $\sqrt{x^2 + 1} \approx x$ when $x \to \infty$.

5. We observe that as $x \to \infty$, the denominator $x + \sqrt{x^2 + 1}$ also tends to infinity: $x + \sqrt{x^2 + 1} \approx x + x = 2x$ So, the limit simplifies to: $\lim_{x \to \infty} \frac{-1}{2x} = 0$

Therefore, $\lim_{x \to \infty} \left( x - \sqrt{x^2 + 1} \right) = 0$

Would you like more details or have any questions about this solution? Here are five more questions you might be interested in:

1. How do you evaluate limits involving square roots?
2. What are the different techniques for evaluating limits at infinity?
3. How do you apply L'Hôpital's rule to solve indeterminate forms?
4. How do you simplify expressions involving conjugates?
5. What are some common limit problems and their solutions?

Tip: When dealing with limits at infinity involving square roots, rationalizing the expression can often simplify the problem and make it easier to evaluate the limit.