To find the left-hand limit of the function $f(x) = \sqrt{x - 6}$ as $x \to 6^-$, let's examine the behavior of the square root function near 6 from the left.

Function and Limit Setup:

We are asked to find:

$\lim_{x \to 6^-} \sqrt{x - 6}$

Behavior of $f(x)$ as $x \to 6^-$:

• The function $f(x) = \sqrt{x - 6}$ is defined only for $x \geq 6$, since the square root of a negative number is not a real number.
• As $x \to 6^-$ (from values less than 6), $x - 6$ becomes negative, which means $\sqrt{x - 6}$ is not defined for real numbers.
• Therefore, the limit does not exist in the real number system.

Conclusion:

$\lim_{x \to 6^-} \sqrt{x - 6} = \text{Does not exist (DNE)}$

Would you like further clarification or additional details?

Here are 5 related questions to consider:

1. What is the right-hand limit of $f(x) = \sqrt{x - 6}$ as $x \to 6^+$?
2. How do we evaluate limits for functions involving square roots near undefined points?
3. What is the domain of the function $f(x) = \sqrt{x - 6}$?
4. Can we find a limit involving $f(x) = \sqrt{x - 6}$ as $x \to \infty$?
5. How do we handle limits of piecewise functions near points of discontinuity?

Tip: Always check the domain of a function before calculating limits to ensure the function is defined near the point of interest.