Chapter 3: Problem 34
In Problems \(29-48\), find the limits. $$ \lim _{x \rightarrow-2} \sqrt{6+x} $$
Short Answer
Expert verified
The limit is 2.
Step by step solution
01
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 \(-2\).
02
Substitute the Limit Point
To find the limit of \( \sqrt{6 + x} \) as \( x \) approaches \(-2\), directly substitute \( x = -2 \) into the expression: \( \sqrt{6 - 2} \).
03
Simplify the Expression
After substitution, simplify the expression: \( \sqrt{6 - 2} = \sqrt{4} \).
04
Calculate the Limit
Calculate \( \sqrt{4} \), which equals 2. Therefore, the limit is 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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 \(\sqrt{6+x}\):
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.
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:
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.