Problem 33
Question
In Problems 29-48, find the limits. $$ \lim _{x \rightarrow-1} \sqrt{4+5 x^{4}} $$
Step-by-Step Solution
Verified Answer
The limit is 3.
1Step 1: Recognize the Function Structure
The given function is \( \sqrt{4 + 5x^4} \). Our task is to find the limit as \( x \to -1 \). Since this is a square root function, we must be careful about domain restrictions, but the expression inside the square root will remain positive for the values we consider.
2Step 2: Substitute the Limit Value
Since the function \( \sqrt{4 + 5x^4} \) is continuous for all real \( x \), we can directly substitute \( x = -1 \) into the expression to find the limit. This is possible because there is no discontinuity or undefined behavior at \( x = -1 \).
3Step 3: Calculate the Expression
Substitute \( x = -1 \) into the expression inside the square root: \[ 4 + 5(-1)^4 = 4 + 5(1) = 4 + 5 = 9 \]Now, take the square root of the result: \( \sqrt{9} = 3 \).
4Step 4: Conclude with the Limit Value
Having substituted \( x = -1 \) and calculated inside the square root, we find \( \sqrt{9} = 3 \). Thus, the limit is \( 3 \).
Key Concepts
Square Root FunctionsContinuity in FunctionsSubstitution Method in Limits
Square Root Functions
Square root functions are a category of functions denoted by the square root symbol \( \sqrt{} \). They have a form \( \sqrt{f(x)} \), where \( f(x) \) is the function inside the root.
These functions are unique because they inherently restrict the domain, meaning the values \( x \) can take. This is because the expression under the square root must be non-negative to produce real number results. However, in the case of our current exercise, the function \( \sqrt{4 + 5x^4} \) remains positive for all real numbers \( x \). That's because \( 5x^4 \) is always non-negative and adding 4 ensures a positive outcome.
Understanding this attribute of square root functions is crucial when analyzing limits, as it ensures the limit can exist at a given point, provided there's no issue with taking a square root of negative values.
These functions are unique because they inherently restrict the domain, meaning the values \( x \) can take. This is because the expression under the square root must be non-negative to produce real number results. However, in the case of our current exercise, the function \( \sqrt{4 + 5x^4} \) remains positive for all real numbers \( x \). That's because \( 5x^4 \) is always non-negative and adding 4 ensures a positive outcome.
Understanding this attribute of square root functions is crucial when analyzing limits, as it ensures the limit can exist at a given point, provided there's no issue with taking a square root of negative values.
Continuity in Functions
Continuity is a core concept in calculus and a crucial property when finding limits. For a function to be continuous at a point \( x = a \), the following needs to be true: the function is defined at \( a \), the limit as \( x \to a \) exists, and the limit equals the function's value at \( a \).
This directly applies to our original problem where we examine the function \( \sqrt{4 + 5x^4} \). Since this function is a composition of continuous functions (polynomials are continuous), it remains continuous over its entire domain. Specifically, it's continuous at \( x = -1 \).
Thanks to its continuity at this point, we can evaluate the limit by simple substitution because there aren't any jumps, holes, or asymptotes near or at \( x = -1 \).
This directly applies to our original problem where we examine the function \( \sqrt{4 + 5x^4} \). Since this function is a composition of continuous functions (polynomials are continuous), it remains continuous over its entire domain. Specifically, it's continuous at \( x = -1 \).
Thanks to its continuity at this point, we can evaluate the limit by simple substitution because there aren't any jumps, holes, or asymptotes near or at \( x = -1 \).
- Check the function is defined at the point.
- Ensure continuity for seamless substitution in limit evaluation.
- Use continuity to simplify finding limits.
Substitution Method in Limits
The substitution method is a powerful technique in finding limits. It applies directly when the function in question is continuous at the point of interest.
In our exercise, we used the substitution method to find the limit at \( x = -1 \) for the function \( \sqrt{4 + 5x^4} \). Since the function is continuous here, as explained, we can substitute \( x = -1 \) directly into \( 4 + 5x^4 \). This step simplifies our work to simple arithmetic instead of more complex techniques.
Here's a brief outline of applying substitution:
In our exercise, we used the substitution method to find the limit at \( x = -1 \) for the function \( \sqrt{4 + 5x^4} \). Since the function is continuous here, as explained, we can substitute \( x = -1 \) directly into \( 4 + 5x^4 \). This step simplifies our work to simple arithmetic instead of more complex techniques.
Here's a brief outline of applying substitution:
- Verify the function's continuity at the given point.
- Substitute the value into the function directly.
- Simplify to find the limit.
Other exercises in this chapter
Problem 32
In Problems 29-48, find the limits. $$ \lim _{x \rightarrow-\pi / 2} \frac{1+\tan ^{2} x}{\sec ^{2} x} $$
View solution Problem 32
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 0} \frac{\sqrt{2-x}-\sqrt{2}}{2 x} $$
View solution Problem 33
Use a table and a graph to find out what happens to $$ f(x)=\frac{2}{x}-\frac{1}{x^{2}} $$ as \(x \rightarrow 0^{-}\). What happens as \(x \rightarrow 0^{+}\) ?
View solution Problem 34
In Problems 29-48, find the limits. $$ \lim _{x \rightarrow-2} \sqrt{6+x} $$
View solution