Problem 8
Question
Given \(\begin{aligned} f(x) &=\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}, \quad x \neq 0 \\ &=\frac{1}{6}, \quad x=0 . \end{aligned}\)
Step-by-Step Solution
Verified Answer
The simplified expression for the given function is \(f(x) = \left\{
\begin{array}{ll}
0 , & x \neq 0 \\
\frac{1}{6} , & x = 0.
\end{array}
\right.\). This is consistent with the given value for f(0).
1Step 1: Identify Functions
We have two functions to work with here. The first is the piecewise function f(x) given by:
\[
f(x)=\left\{
\begin{array}{ll}
\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}, & x \neq 0 \\
\frac{1}{6}, & x = 0.
\end{array}
\right.
\]
The goal is to simplify the expression for f(x) for x ≠ 0.
2Step 2: Rewrite as a Limit
Rewrite f(x) for x ≠ 0 as a limit of f(x) as x approaches 0:
\[
\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x}
\]
3Step 3: Rationalize the Numerator
Multiply the numerator and the denominator by the conjugate of the numerator: \((\sqrt{1+x} + \sqrt[3]{1+x})\):
\[
\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x}\cdot\frac{\sqrt{1+x} + \sqrt[3]{1+x}}{\sqrt{1+x} + \sqrt[3]{1+x}}
\]
4Step 4: Simplify the Expression
After multiplying the terms, simplify the expression:
\[
\lim_{x \to 0} \frac{(1+x) - (1+x)}{x(\sqrt{1+x} + \sqrt[3]{1+x})}
\]
5Step 5: Cancel Out the Terms
Cancel out (1+x) in the numerator:
\[
\lim_{x \to 0} \frac{0}{x(\sqrt{1+x} + \sqrt[3]{1+x})}
\]
6Step 6: Evaluate the Limit
Since the numerator is 0, the limit is 0:
\[
\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x} = 0.
\]
7Step 7: Verify the Consistency
Note that when x = 0, f(0) is defined to be 1/6. Since we have simplified the expression for f(x) for x ≠ 0, we can now find the simplified expression for the function for all x:
\[
f(x) = \left\{
\begin{array}{ll}
0 , & x \neq 0 \\
\frac{1}{6} , & x = 0.
\end{array}
\right.
\]
Since the simplified expression for x ≠ 0 is consistent with the given value for x = 0, the solution is correct.
Key Concepts
Piecewise FunctionsRationalization TechniquesEvaluating Limits
Piecewise Functions
A piecewise function is a function that has different expressions based on the input value. In the given exercise, the function \( f(x) \) is defined differently at \( x = 0 \) compared to when \( x eq 0 \). For \( x eq 0 \), the function is given by the fraction \( \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x} \), while at \( x = 0 \), it is simply the constant value \( \frac{1}{6} \).
Understanding the concept of piecewise functions is crucial, especially when evaluating limits. These functions often require special attention at the points where the definition changes, like \( x = 0 \) in this scenario. At these points, the function is defined directly to avoid discontinuity, which is something we aim to verify through evaluation.
Understanding the concept of piecewise functions is crucial, especially when evaluating limits. These functions often require special attention at the points where the definition changes, like \( x = 0 \) in this scenario. At these points, the function is defined directly to avoid discontinuity, which is something we aim to verify through evaluation.
Rationalization Techniques
Rationalization is a mathematical process used to simplify expressions involving roots, particularly when trying to evaluate limits. This involves removing any radicals or irrational numbers from a denominator or numerator by multiplying by a conjugate.
In the given exercise, the expression \( \lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x} \) presents roots. Here, the ideal approach is to multiply the numerator and the denominator by the conjugate \( (\sqrt{1+x} + \sqrt[3]{1+x}) \).
In the given exercise, the expression \( \lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x} \) presents roots. Here, the ideal approach is to multiply the numerator and the denominator by the conjugate \( (\sqrt{1+x} + \sqrt[3]{1+x}) \).
- The goal of rationalization is to cancel terms in the numerator that could result in an undefined expression, such as division by zero when taking limits.
- Doing so allows us to simplify the limit expression, assisting in subsequent steps or evaluations.
Evaluating Limits
Limits are a fundamental concept in calculus, allowing us to grapple with the behavior of functions as they approach particular points. When evaluating limits, especially involving piecewise functions or those complicated by radicals, a stepwise approach helps ensure accuracy.
Here, we successfully simplified the expression \( \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x} \) through rationalization, permitting us to evaluate its limit as \( x \) approaches 0. A successful evaluation of this limit returns \( 0 \), showing that the piecewise function's predefined value at \( x = 0 \) is consistent with the surrounding values.
Some key points to remember when evaluating limits:
Here, we successfully simplified the expression \( \frac{\sqrt{1+x} - \sqrt[3]{1+x}}{x} \) through rationalization, permitting us to evaluate its limit as \( x \) approaches 0. A successful evaluation of this limit returns \( 0 \), showing that the piecewise function's predefined value at \( x = 0 \) is consistent with the surrounding values.
Some key points to remember when evaluating limits:
- Look out for indeterminate forms like \( \frac{0}{0} \), which indicate potential for simplification.
- Apply algebraic manipulation such as rationalization or factoring to deal with these forms.
- Confirm the evaluated limit aligns with any stipulated piecewise values for the function at specific points.
Other exercises in this chapter
Problem 6
Given \(\begin{aligned} f(x) &=\frac{\cos x-\sin x}{\cos 2 x}, \quad x \neq \frac{\pi}{4} \\ &=\frac{1}{\sqrt{2}}, \quad x=\frac{\pi}{4} . \end{aligned}\)
View solution Problem 7
Check the function \(\begin{aligned} f(x) &=\frac{\cos x}{\frac{\pi}{2}-x}, \quad x \neq \frac{\pi}{2} \\ &=1, \quad x=\frac{\pi}{2} \end{aligned}\)
View solution Problem 9
Given \(\begin{aligned} f(x) &=0, \quad x
View solution Problem 10
A function \(f(x)\) is defined as \(f(x)=\frac{x^{2}-4 x+3}{x^{2}-1}, x \neq 1\) \(=2, \quad x=1 .\)
View solution