Problem 38
Question
Limits of quotients Find the limits in Exercises \(23-42\) $$\lim _{x \rightarrow-1} \frac{\sqrt{x^{2}+8}-3}{x+1}$$
Step-by-Step Solution
Verified Answer
The limit is \(-\frac{1}{3}\).
1Step 1: Identify the Limit Expression
We need to find the limit \( \lim _{x \rightarrow -1} \frac{\sqrt{x^{2}+8}-3}{x+1} \). The presence of \( x \rightarrow -1 \) indicates we are substituting \( x = -1 \) into the function, but doing so directly results in a \(\frac{0}{0}\) indeterminate form.
2Step 2: Multiply by the Conjugate
Multiply the numerator and the denominator by the conjugate of the numerator: \( \frac{\sqrt{x^2 + 8} - 3}{x + 1} \times \frac{\sqrt{x^2 + 8} + 3}{\sqrt{x^2 + 8} + 3} \). This helps to eliminate the square root in the numerator.
3Step 3: Simplify the Expression
The numerator becomes \((\sqrt{x^2 + 8})^2 - 3^2 = x^2 + 8 - 9 = x^2 - 1\). The expression is now \(\frac{x^2 - 1}{(x + 1)( \sqrt{x^2 + 8} + 3)}\).
4Step 4: Factor the Numerator
Factor \(x^2 - 1\) as \((x - 1)(x + 1)\). The expression becomes \(\frac{(x - 1)(x + 1)}{(x + 1)(\sqrt{x^2 + 8} + 3)}\).
5Step 5: Cancel Common Terms
Cancel the common term \((x + 1)\) from the numerator and the denominator. The simplified expression is \(\frac{x - 1}{\sqrt{x^2 + 8} + 3}\).
6Step 6: Evaluate the Limit
Substitute \(x = -1\) into the simplified expression: \(\frac{-1 - 1}{\sqrt{(-1)^2 + 8} + 3} = \frac{-2}{\sqrt{1 + 8} + 3} = \frac{-2}{\sqrt{9} + 3} = \frac{-2}{3 + 3} = \frac{-2}{6} = -\frac{1}{3}\).
Key Concepts
Indeterminate formsAlgebraic manipulationLimit evaluation techniques
Indeterminate forms
In calculus, an indeterminate form often arises when evaluating limits. Specifically, when we substitute a certain value into a function and the result is an expression like \( \frac{0}{0} \), it signals an indeterminate form. These forms suggest that the function behaves unpredictably near the point of interest. This means that a direct substitution won't provide a valid answer.
To handle such cases, we must manipulate the expression creatively to resolve the indeterminacy. In this exercise, substituting \( x = -1 \) into \( \lim_{x \rightarrow -1} \frac{\sqrt{x^{2}+8}-3}{x+1} \) results in \( \frac{0}{0} \). Therefore, we must apply additional techniques, such as algebraic manipulation or using limits properties, to evaluate the expression correctly and determine the limit.
To handle such cases, we must manipulate the expression creatively to resolve the indeterminacy. In this exercise, substituting \( x = -1 \) into \( \lim_{x \rightarrow -1} \frac{\sqrt{x^{2}+8}-3}{x+1} \) results in \( \frac{0}{0} \). Therefore, we must apply additional techniques, such as algebraic manipulation or using limits properties, to evaluate the expression correctly and determine the limit.
Algebraic manipulation
Algebraic manipulation is a critical tool in resolving indeterminate forms. These techniques involve rearranging or transforming the expression to make it easier to evaluate the limit. For this problem, we used the method of multiplying by the conjugate.
The original expression \( \frac{\sqrt{x^{2}+8}-3}{x+1} \) was multiplied by \( \frac{\sqrt{x^2 + 8} + 3}{\sqrt{x^2 + 8} + 3} \). This operation helps remove the square root from the numerator by taking advantage of the difference of squares identity: \( (a-b)(a+b) = a^2 - b^2 \).
After simplification, the expression changes to \( \frac{x^2 - 1}{(x + 1)( \sqrt{x^2 + 8} + 3)} \). Further simplification involves factoring the numerator \( x^2 - 1 \) as \( (x-1)(x+1) \). As a result, you can cancel the common term \( (x + 1) \) in the numerator and denominator, simplifying the expression to \( \frac{x - 1}{\sqrt{x^2 + 8} + 3} \). This manipulation process is essential to help evaluate the limit accurately.
The original expression \( \frac{\sqrt{x^{2}+8}-3}{x+1} \) was multiplied by \( \frac{\sqrt{x^2 + 8} + 3}{\sqrt{x^2 + 8} + 3} \). This operation helps remove the square root from the numerator by taking advantage of the difference of squares identity: \( (a-b)(a+b) = a^2 - b^2 \).
After simplification, the expression changes to \( \frac{x^2 - 1}{(x + 1)( \sqrt{x^2 + 8} + 3)} \). Further simplification involves factoring the numerator \( x^2 - 1 \) as \( (x-1)(x+1) \). As a result, you can cancel the common term \( (x + 1) \) in the numerator and denominator, simplifying the expression to \( \frac{x - 1}{\sqrt{x^2 + 8} + 3} \). This manipulation process is essential to help evaluate the limit accurately.
Limit evaluation techniques
Evaluating limits accurately requires specific techniques, especially when dealing with complex expressions. Common techniques include direct substitution, algebraic simplification, and in some cases, L'Hôpital's Rule. In our case, once the expression is manipulated to a simpler form, direct substitution can be applied effectively.
After simplifying the expression to \( \frac{x - 1}{\sqrt{x^2 + 8} + 3} \), we substitute \( x = -1 \) directly. This yields \( \frac{-2}{\sqrt{1 + 8} + 3} \). By simplifying further: \( \sqrt{9} = 3 \), the expression becomes \( \frac{-2}{3 + 3} = \frac{-2}{6} = -\frac{1}{3} \).
In summary, applying these limit evaluation techniques enables us to tackle the complex nature of indeterminate forms and arrive at accurate results. Knowing when and how to utilize these techniques is crucial for mastering limits in calculus.
After simplifying the expression to \( \frac{x - 1}{\sqrt{x^2 + 8} + 3} \), we substitute \( x = -1 \) directly. This yields \( \frac{-2}{\sqrt{1 + 8} + 3} \). By simplifying further: \( \sqrt{9} = 3 \), the expression becomes \( \frac{-2}{3 + 3} = \frac{-2}{6} = -\frac{1}{3} \).
In summary, applying these limit evaluation techniques enables us to tackle the complex nature of indeterminate forms and arrive at accurate results. Knowing when and how to utilize these techniques is crucial for mastering limits in calculus.
Other exercises in this chapter
Problem 38
Find the limits in Exercises \(37-48.\) $$\lim _{x \rightarrow 0^{-}} \frac{5}{2 x}$$
View solution Problem 38
Find the limits in Exercises \(21-42\) $$\lim _{\theta \rightarrow 0} \sin \theta \cot 2 \theta$$
View solution Problem 39
Define \(f(1)\) in a way that extends \(f(s)=\left(s^{3}-1\right) /\left(s^{2}-1\right)\) to be continuous at \(s=1\)
View solution Problem 39
Find the limits in Exercises \(37-48.\) $$\lim _{x \rightarrow 2^{-}} \frac{3}{x-2}$$
View solution