To solve the limit problem $\lim_{{x \to -\infty}} \left( x^2 + x \sqrt{x^2 - 1} \right)$ using the method of multiplying by the conjugate, we can proceed as follows:

Given: $\lim_{{x \to -\infty}} \left( x^2 + x \sqrt{x^2 - 1} \right)$

Rewrite the expression by factoring out $x^2$ inside the square root: $= \lim_{{x \to -\infty}} \left( x^2 + x \sqrt{x^2 \left( 1 - \frac{1}{x^2} \right)} \right)$

Since $\sqrt{x^2 \left( 1 - \frac{1}{x^2} \right)} = |x| \sqrt{1 - \frac{1}{x^2}}$, and for $x \to -\infty$, $|x| = -x$, we get: $= \lim_{{x \to -\infty}} \left( x^2 + x(-x) \sqrt{1 - \frac{1}{x^2}} \right)$ $= \lim_{{x \to -\infty}} \left( x^2 - x^2 \sqrt{1 - \frac{1}{x^2}} \right)$ $= \lim_{{x \to -\infty}} x^2 \left( 1 - \sqrt{1 - \frac{1}{x^2}} \right)$

Now, multiply by the conjugate: $= \lim_{{x \to -\infty}} x^2 \cdot \frac{\left( 1 - \sqrt{1 - \frac{1}{x^2}} \right) \left( 1 + \sqrt{1 - \frac{1}{x^2}} \right)}{1 + \sqrt{1 - \frac{1}{x^2}}}$ $= \lim_{{x \to -\infty}} x^2 \cdot \frac{1 - \left(1 - \frac{1}{x^2}\right)}{1 + \sqrt{1 - \frac{1}{x^2}}}$ $= \lim_{{x \to -\infty}} x^2 \cdot \frac{\frac{1}{x^2}}{1 + \sqrt{1 - \frac{1}{x^2}}}$ $= \lim_{{x \to -\infty}} \frac{1}{1 + \sqrt{1 - \frac{1}{x^2}}}$

As $x \to -\infty$, $\frac{1}{x^2} \to 0$, so the expression becomes: $= \frac{1}{1 + 1} = \frac{1}{2}$

Thus, the answer is: $\boxed{\frac{1}{2}}$

Would you like further details on each step, or do you have any other questions?

Related Questions:

1. What happens if $x$ approaches $+\infty$ instead of $-\infty$?
2. How do we handle limits at infinity in more complex radical expressions?
3. Can you explain why multiplying by the conjugate is helpful in this problem?
4. What would happen if there were additional terms inside the square root?
5. Are there alternative methods to solve this limit besides conjugate multiplication?

Tip:

When dealing with limits involving square roots at infinity, try simplifying under the square root and considering conjugate multiplication to eliminate indeterminate forms.