Problem 59 - Precalculus with Limits

Question

In Exercises 49-68, find the limit by direct substitution. $$ \lim_{x \to -1}\ \sqrt{x+2}$$

Step-by-Step Solution

Answer

The limit of \( \sqrt{x+2} \) as \( x \) approaches -1 is 1.

Step 1: Understand the function

The function given in this case is \( \sqrt{x+2} \), a square root function. It is defined for all \( x \) such that \( x >= -2 \).

Step 2: Direct substitution

To find a limit using direct substitution, simply replace \( x \) with the value it is approaching, which here is -1. Substituting \( x = -1 \) into \( \sqrt{x+2} \) leads to \( \sqrt{-1+2} = \sqrt{1} \).

Step 3: Finalize the limit

Since the result of the substitution is a real number, and is not an indeterminate form, this means the limit of the function as \( x \) approaches -1 equals to the result, which is \(\sqrt{1}=1\).

Key Concepts

Direct Substitution

Direct substitution is one of the simplest methods for finding the limit of a function as the variable approaches a certain value. The idea is straightforward: you replace the variable with the value it is approaching. If the result is a real number, then that is the limit of the function. For example, consider the function given in the exercise, \( \sqrt{x+2} \). To find the limit as \( x \to -1 \), you substitute \( -1 \) in place of \( x \). This means calculating \( \sqrt{-1+2} \), which simplifies to \( \sqrt{1} \), yielding a final result of 1.

When applying direct substitution, always check that the function is defined at the point you're evaluating. In this case, since the function \( \sqrt{x+2} \) is defined for \( x \ge -2 \), direct substitution is valid. If a function results in an undefined form or a complex number, other methods may be necessary.

Square Root Function

Square root functions, like \( \sqrt{x+2} \), are continuous and smooth for all values within their domain. Typically, the square root symbol \( \sqrt{} \) implies only the positive root. The function \( \sqrt{x+2} \) is well-defined for values of \( x \) such that \( x \ge -2 \), because you cannot take the square root of a negative number in real numbers.

Key properties include:

• Continuous within their domain.
• Non-negative output for non-negative input under real numbers.
• As \( x \) approaches the edge of the domain (i.e., \(-2\) in this scenario), the function approaches zero.

For our specific limit in the exercise, since -1 is well within the domain, the function behaves predictably and yields a simple result of \( \sqrt{1} = 1 \). Understanding these characteristics helps you know when direct substitution will work effectively in limit problems.

Indeterminate Forms

Indeterminate forms are situations in calculus where the limit cannot be directly determined by simple substitution. Common forms include \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). These forms require special techniques like L'Hôpital's rule or algebraic manipulation to resolve and find the limit.

In the provided exercise, when using direct substitution to evaluate \( \lim_{x \to -1} \sqrt{x+2} \), the result is a clear number, \( \sqrt{1} = 1 \). This indicates no indeterminate form is present here, as the substitution results in a defined and real value. Recognizing when an expression is and isn’t indeterminate is crucial. It allows you to decide whether direct substitution will suffice or if a more complex method is needed.