Problem 20 - Precalculus

Question

Evaluate the following limits using direct substitution, if possible. If not possible, state why. $$\lim _{x \rightarrow-6} \sqrt{x^{2}+7 x}$$

Step-by-Step Solution

Answer

Limit does not exist due to the square root of a negative number.

Step 1: Check if Direct Substitution is Possible

Firstly, evaluate the expression within the square root when \(x = -6\). We substitute \(-6\) into \(x^2 + 7x\) to get \((-6)^2 + 7(-6)\).

Step 2: Calculate the Expression Inside the Square Root

Calculate \((-6)^2 + 7(-6)\). This expands to \(36 - 42\), which equals \(-6\).

Step 3: Determine Feasibility of Direct Substitution

Since the expression inside the square root, \(-6\), is negative, the square root of a negative number is not defined in the real number system. Thus, direct substitution is not possible.

Step 4: State the Conclusion

Because direct substitution leads to the square root of a negative number, and we are considering real-valued limits, this limit does not exist.

Key Concepts

Direct Substitution

Direct substitution is one of the simplest techniques for evaluating limits in calculus. In this method, we directly replace the variable with the value it approaches. Here, the key is to check if the function remains defined after substitution. For instance, in the given exercise, we substitute \(x = -6\) into the expression within the square root. This step determines whether the limit can be evaluated straightforwardly or if further analysis is necessary. Direct substitution works well when the function is continuous and defined at the point of interest. However, if it leads to an undefined form, further methods must be employed.

Square Root Functions

Square root functions are quite common in calculus and have unique characteristics. They only produce real numbers for non-negative inputs. This feature is because, in the real number system, taking the square root of a negative number does not yield a real number.
For example, in \(f(x) = \sqrt{x}\), the function is defined only for \(x \geq 0\). When a square root function appears in a limit problem, we must ensure that the expression inside the square root is non-negative.
This is exactly why, in our example, the limit does not exist using direct substitution. After substitution, the expression inside the square root \((-6)\) is negative, making the square root undefined in the real number system.

Real Number System

The real number system is a crucial concept in calculus, especially when dealing with functions and limits. It consists of all rational and irrational numbers that can be plotted on a number line. Negative numbers do exist in this system, but limits involving these require careful attention when plugged into a context where certain operations, like square rooting, aren't defined.
When evaluating limits, if the outcome of operations results in the square root of a negative number, it falls outside the regular real number system. Thus, we must either consider complex numbers (not typically in a standard calculus course) or state that the limit is undefined, just as it happened in our example.

Evaluating Limits

Evaluating limits is at the heart of calculus, helping us understand the behavior of functions as they approach specific points. A rigorous understanding clears the path for grasping continuity, derivatives, and much more.
In our example, checking if direct substitution is possible was the initial step in evaluating the limit. We substituted the value directly, and the outcome was a negative square root, leading to the conclusion that the limit does not exist under real numbers.

• Always substitute the approaching value first, if possible.
• If results are undefined or complex in nature within real numbers, further investigation is needed.
• For any square root result, ensure the rooted value is non-negative in the set of real numbers.

These steps ensure a structured framework when evaluating limits while adhering to the properties of the real number system.