Problem 24
Question
Apply l'Hôpital's Rule repeatedly (when needed) to evaluate the given limit, if it exists. \(\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}-2 x}{\sin \left(x^{3}\right)}\)
Step-by-Step Solution
Verified Answer
The limit is \(\frac{1}{3}\).
1Step 1: Check the Form of the Limit
First, substitute \(x = 0\) into the expression \(\frac{e^{x}-e^{-x}-2 x}{\sin(x^3)}\). This results in \(\frac{e^{0}-e^{0}-2 \cdot 0}{\sin(0^3)} = \frac{0}{0}\), which is an indeterminate form. Therefore, we can apply l'Hôpital's Rule.
2Step 2: Apply l'Hôpital's Rule
Since the initial substitution leads to \(\frac{0}{0}\), differentiate the numerator and denominator with respect to \(x\):- The derivative of the numerator \(e^x - e^{-x} - 2x\) is \(e^x + e^{-x} - 2\).- The derivative of the denominator \(\sin(x^3)\) is \(3x^2 \cos(x^3)\).So, the limit becomes \(\lim_{x \to 0} \frac{e^x + e^{-x} - 2}{3x^2 \cos(x^3)}\). Substitute \(x = 0\) to check the new form: \(\frac{0}{0}\). Another indeterminate form, applying l'Hôpital's Rule again is necessary.
3Step 3: Apply l'Hôpital's Rule Again
Differentiate the new numerator \(e^x + e^{-x} - 2\), resulting in \(e^x - e^{-x}\). Differentiate the new denominator \(3x^2 \cos(x^3)\), resulting in \(6x \cos(x^3) - 9x^5 \sin(x^3)\) using the product rule.The limit is now \(\lim_{x \to 0} \frac{e^x - e^{-x}}{6x \cos(x^3) - 9x^5 \sin(x^3)}\). Substitute \(x = 0\) again: \(\frac{0}{0}\), still indeterminate. Apply l'Hôpital's Rule one more time.
4Step 4: Apply l'Hôpital's Rule One Final Time
Differentiate \(e^x - e^{-x}\) to get \(e^x + e^{-x}\). Differentiate \(6x \cos(x^3) - 9x^5 \sin(x^3)\):- This becomes \([6 \cos(x^3) - 18x^3 \sin(x^3)] + [-27x^4 \cos(x^3)]\).Thus, the expression changes to \(\lim_{x \to 0} \frac{e^x + e^{-x}}{6 \cos(x^3) - 18x^3 \sin(x^3) - 27x^4 \cos(x^3)}\). Substitute \(x = 0\), obtaining \(\frac{2}{6} = \frac{1}{3}\), which is a determinate form.
Key Concepts
Indeterminate FormsLimitsDifferentiation
Indeterminate Forms
In calculus, you often encounter expressions that result in forms which are undefined at first glance, known as indeterminate forms. These are typically forms that emerge when attempting to analyze limits, such as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), and others.
This indeterminate nature means that further work is necessary to see if the limit exists and, if so, what value it may take. When we evaluated the given limit initially, substituting \( x = 0 \) gave us \( \frac{0}{0} \), a classic indeterminate form.
This was a key indicator to make use of l'Hôpital's Rule, which is specifically designed to resolve such ambiguities by applying differentiation.
This indeterminate nature means that further work is necessary to see if the limit exists and, if so, what value it may take. When we evaluated the given limit initially, substituting \( x = 0 \) gave us \( \frac{0}{0} \), a classic indeterminate form.
This was a key indicator to make use of l'Hôpital's Rule, which is specifically designed to resolve such ambiguities by applying differentiation.
Limits
Limits are fundamental to understanding calculus. They help us understand the values that functions approach as the input approaches a certain point. Calculating limits allows mathematicians to bridge the gap between discrete and continuous mathematics.
In our exercise, the limit is evaluated as \( x \to 0 \). This means we are seeking the value that the expression "wants" to reach as \( x \) gets closer and closer to 0. However, to determine this, we must delve into processes like differentiation, especially when indeterminate forms present themselves.
After multiple applications of l'Hôpital's Rule by differentiating the function's numerator and denominator, we revealed that the limit simplifies to \( \frac{1}{3} \), a definitive value. This illustrates how limits can sometimes require intricate steps such as repeated differentiation to uncover their true behavior.
In our exercise, the limit is evaluated as \( x \to 0 \). This means we are seeking the value that the expression "wants" to reach as \( x \) gets closer and closer to 0. However, to determine this, we must delve into processes like differentiation, especially when indeterminate forms present themselves.
After multiple applications of l'Hôpital's Rule by differentiating the function's numerator and denominator, we revealed that the limit simplifies to \( \frac{1}{3} \), a definitive value. This illustrates how limits can sometimes require intricate steps such as repeated differentiation to uncover their true behavior.
Differentiation
Differentiation is a process of finding the derivative, which measures how a function changes as its input changes. The derivative offers information about the slope or rate of change of the function at any given point and is a crucial tool in calculus.
Applying differentiation was essential in solving this problem using l'Hôpital's Rule. When faced with the initial indeterminate form \( \frac{0}{0} \), we needed to differentiate the numerator \( e^x - e^{-x} - 2x \) and the denominator \( \sin(x^3) \).
The differentiation resulted in new expressions like \( e^x + e^{-x} - 2 \) for the numerator, and \( 3x^2 \cos(x^3) \) for the denominator. Continued differentiation further simplified the limit each time it reappeared as an indeterminate form. By performing these steps repeatedly, we transformed a problematic fraction into a clear solution, ultimately determining the limit as \( \frac{1}{3} \), demonstrating the power of differentiation to illuminate seemingly complex problems.
Applying differentiation was essential in solving this problem using l'Hôpital's Rule. When faced with the initial indeterminate form \( \frac{0}{0} \), we needed to differentiate the numerator \( e^x - e^{-x} - 2x \) and the denominator \( \sin(x^3) \).
The differentiation resulted in new expressions like \( e^x + e^{-x} - 2 \) for the numerator, and \( 3x^2 \cos(x^3) \) for the denominator. Continued differentiation further simplified the limit each time it reappeared as an indeterminate form. By performing these steps repeatedly, we transformed a problematic fraction into a clear solution, ultimately determining the limit as \( \frac{1}{3} \), demonstrating the power of differentiation to illuminate seemingly complex problems.
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