Problem 34 - Calculus for Biology and Medicine

Question

In Problems \(29-48\), find the limits. $$ \lim _{x \rightarrow-2} \sqrt{6+x} $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Identify the Expression

The expression in the problem is \( \sqrt{6 + x} \). Our task is to find the limit of this expression as \( x \) approaches \(-2\).

Step 2: Substitute the Limit Point

To find the limit of \( \sqrt{6 + x} \) as \( x \) approaches \(-2\), directly substitute \( x = -2 \) into the expression: \( \sqrt{6 - 2} \).

Step 3: Simplify the Expression

After substitution, simplify the expression: \( \sqrt{6 - 2} = \sqrt{4} \).

Step 4: Calculate the Limit

Calculate \( \sqrt{4} \), which equals 2. Therefore, the limit is 2.

Key Concepts

Understanding Radical Expressions

Radical expressions are mathematical phrases that include a radical sign, which looks like a check mark with a horizontal line on top. The most common radical expression is the square root, denoted as \(\sqrt{}\). In calculus, you'll often encounter problems involving radicals when finding limits. These expressions can sometimes look intimidating, but they're simply asking you to perform the operation of finding the root of a number or expression.
Let’s break it down further:

• The number inside the radical is called the radicand.
• For example, in \(\sqrt{6 + x}\), \(6 + x\) is the radicand.
• The root symbol denotes which root you should take; the square root is the second root.

Understanding how to manipulate and simplify these expressions is crucial. In our exercise, knowing how to handle the radical helps simplify finding the limit as \(x\) approaches \(-2\). This is all about recognizing how changing values will affect the result under the radical sign.

The Direct Substitution Method

The direct substitution method is a straightforward way to find limits in calculus. This approach works perfectly when the function is continuous at the point that \(x\) is approaching.
Here's how it works when using a radical expression like \(\sqrt{6+x}\):

• First, take the value that \(x\) is approaching, in this case, \(-2\).
• Substitute this value directly into the expression: \(\sqrt{6 + (-2)}\).
• Then, simplify it down to find \(\sqrt{4}\), which equals 2.

This method is incredibly convenient since it involves minimal computation if the function is well-behaved and doesn't hit any undefined points, such as division by zero. In our problem, the direct substitution method readily gives us the limit by taking advantage of the simplicity of radical expressions.

Exploring Continuity of Functions

Continuity is a fundamental concept in calculus that refers to a function's behavior over its domain. A function is continuous at a point if you can draw it without lifting your pencil from the paper. In our problem, the function \(\sqrt{6+x}\) is continuous over its entire domain.
Here's what makes a function like \(\sqrt{6+x}\) continuous:

• The function is defined at the point \(x = -2\) since substituting \(-2\) results in a real number \(\sqrt{4}\).
• There are no breaks, holes, or vertical asymptotes within its domain around \(x = -2\).
• As \(x\) approaches \(-2\) from either the left or the right, the limits are the same, confirming its continuity at that point.

Understanding this concept helps determine when you can apply methods like direct substitution. If a function is not continuous at a point, you'll need to use other techniques to find the limit, such as factoring or rationalizing. The continuity guarantees that approaches like direct substitution will work smoothly for radical functions like the one in our exercise.