Problem 36
Question
In Problems \(29-48\), find the limits. $$ \lim _{x \rightarrow 1} \sqrt{x^{3}+4 x-1} $$
Step-by-Step Solution
Verified Answer
The limit is 2.
1Step 1: Analyze the Limit Expression
The given expression is \( \lim_{x \rightarrow 1} \sqrt{x^3 + 4x - 1} \). We observe it's a limit involving a square root, which can often be directly evaluated at the point if no undefined expressions (like division by zero) or discontinuities occur.
2Step 2: Direct Substitution
Since the expression inside the square root, \( x^3 + 4x - 1 \), is a polynomial, it's continuous over all real numbers. This allows us to use direct substitution to find the limit. Substitute \( x = 1 \) into the expression: \( 1^3 + 4(1) - 1 = 1 + 4 - 1 = 4 \).
3Step 3: Evaluate the Square Root
After substituting \( x = 1 \), the expression inside the limit simplifies to \( \sqrt{4} \). Calculate \( \sqrt{4} = 2 \).
Key Concepts
Polynomial ContinuityDirect SubstitutionSquare Root Evaluation
Polynomial Continuity
When we talk about polynomial continuity, we think about simple, smooth curves on a graph. Polynomials are mathematical expressions involving terms like \( x^2 \), \( x^3 \), or more generally \( x^n \) where \( n \) is a non-negative integer. A great feature of polynomials is their continuity. This means that they don't have any gaps, jumps, or holes when graphed. They stretch smoothly from negative to positive infinity, making them continuous everywhere on the real number line.
For example, the polynomial \( x^3 + 4x - 1 \) in our original exercise is continuous for all values of \( x \). As a result, we can evaluate limits easily by simply substituting the point of interest into the polynomial. Continuity ensures that no abrupt changes occur in the output as we make small changes to the input. This is why polynomial functions are favored for direct substitution in limit evaluation.
For example, the polynomial \( x^3 + 4x - 1 \) in our original exercise is continuous for all values of \( x \). As a result, we can evaluate limits easily by simply substituting the point of interest into the polynomial. Continuity ensures that no abrupt changes occur in the output as we make small changes to the input. This is why polynomial functions are favored for direct substitution in limit evaluation.
Direct Substitution
Direct substitution is a powerful and straightforward method for evaluating limits. It works perfectly when the function involved is continuous at a given point, meaning there's no division by zero or undefined behavior.
In our exercise, we are given the function \( \sqrt{x^3 + 4x - 1} \). First, we find that the expression inside the square root, \( x^3 + 4x - 1 \), is a polynomial, and therefore continuous everywhere. This allows us to apply direct substitution to evaluate the limit as \( x \) approaches 1. By substituting \( x = 1 \) into the polynomial, we simplify it to 4.
Using direct substitution saves effort over an intricate process and confirms whether the function inside the limit can be calculated directly. It's a reliable go-to technique when dealing with polynomials or other well-behaved functions.
In our exercise, we are given the function \( \sqrt{x^3 + 4x - 1} \). First, we find that the expression inside the square root, \( x^3 + 4x - 1 \), is a polynomial, and therefore continuous everywhere. This allows us to apply direct substitution to evaluate the limit as \( x \) approaches 1. By substituting \( x = 1 \) into the polynomial, we simplify it to 4.
Using direct substitution saves effort over an intricate process and confirms whether the function inside the limit can be calculated directly. It's a reliable go-to technique when dealing with polynomials or other well-behaved functions.
Square Root Evaluation
Once direct substitution is applied, what's left is evaluating any remaining expression. In this case, the simplified expression \( \sqrt{4} \) needs assessment.
Square root evaluation involves recognizing that finding the square root is determining a number which, when multiplied by itself, gives the original number under the square root symbol. Here we calculate \( \sqrt{4} \), which is 2 since \( 2 imes 2 = 4 \).
Understanding and evaluating square roots is crucial as it appears in numerous math problems involving limits, calculus, and beyond. Knowing how to evaluate them quickly can make solving complex expressions much easier and more intuitive.
Square root evaluation involves recognizing that finding the square root is determining a number which, when multiplied by itself, gives the original number under the square root symbol. Here we calculate \( \sqrt{4} \), which is 2 since \( 2 imes 2 = 4 \).
Understanding and evaluating square roots is crucial as it appears in numerous math problems involving limits, calculus, and beyond. Knowing how to evaluate them quickly can make solving complex expressions much easier and more intuitive.
Other exercises in this chapter
Problem 35
Use a graphing calculator to investigate $$ \lim _{x \rightarrow 1} \sin \frac{1}{x-1} $$
View solution Problem 35
In Problems \(29-48\), find the limits. $$ \lim _{x \rightarrow-1} \sqrt{x^{2}+2 x+2} $$
View solution Problem 37
In Problems \(37-54\), use the limit laws to evaluate each limit. $$ \lim _{x \rightarrow-1}\left(x^{3}+7 x-1\right) $$
View solution Problem 37
In Problems \(29-48\), find the limits. $$ \lim _{x \rightarrow 0} e^{-x^{2} / 3} $$
View solution