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Think About It In Exercises \(79-82,\) L'Hopital's Rule is used incorrectly. Describe the error. $$\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{e^{x}}=\lim _{x \rightarrow 0} \frac{2 e^{2 x}}{e^{x}}$$ $$=\lim _{x \rightarrow 0} 2 e^{x} \\\=2$$

Short Answer

Expert verified
L'Hopital's Rule was unnecessarily and incorrectly used here because the limit was not in an indeterminate form. The correct answer is 0.

Step by step solution

01

Identify the Misused Rule

Given the problem, the first step is to understand which rule is being applied incorrectly. In this exercise, L'Hopital's Rule is used inaccurately. The condition to apply L'Hopital's Rule is having either '0/0' or '∞/∞' form, when x tends to given limit.
02

Identify the Error

The function \(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{e^{x}}\) does not fall under the '0/0' or '∞/∞' form when x tends to 0. If we substitute x=0 in the limits, we find '0/1’ which is not an indeterminate form. Therefore, L'Hopital's Rule should not have been applied here.
03

Correction and Conclusion

Rather than applying L'Hopital's Rule, the limits should be computed directly which will give \(\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{e^{x}} = \frac{e^0 - 1}{e^0} = \frac{1 - 1}{1} = 0\). The error here was performing an operation (L'Hopital's Rule) on a situation where that operation was not appropriate or needed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Indeterminate Forms
Indeterminate forms often confuse students when dealing with limits. They occur when substituting a value in a limit expression results in an undefined scenario, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These situations require further analysis to find the limit's value. L'Hopital's Rule is a useful technique for solving indeterminate forms. When an expression evaluates to an indeterminate form, L'Hopital's Rule can be applied by differentiating the numerator and the denominator. However, it's crucial to first confirm that the expression indeed presents an indeterminate form before applying L'Hopital's Rule. If a function does not display the form of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hopital's Rule is not applicable.
Techniques in Limit Calculation
Calculating limits is an essential part of calculus, useful for finding the behavior of a function as it approaches a certain point. For expressions that are not obviously indeterminate, like a straightforward division, finding the limit directly may be possible. In the given exercise, substituting \( x = 0 \) in \( \lim_{x \rightarrow 0} \frac{e^{2x} - 1}{e^x} \) directly shows that the expression simplifies to \( \frac{0}{1} \), which equals 0. Instead of incorrectly applying L'Hopital's Rule here, a simple substitution can efficiently solve the expression without complications. Using algebraic simplifications and substitutions are often effective techniques for managing limits where L'Hopital's Rule is inapplicable.
Avoiding Common Calculus Errors
Mistakes in calculus often arise from misunderstanding when specific rules or theorems apply. A common error is using L'Hopital's Rule when it's not warranted. For example, it should only be used when an expression evaluates to an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).To avoid errors:
  • Confirm the form of the limit before using L'Hopital's Rule.
  • Consider direct substitution or algebraic manipulation first.
  • Double-check your work by confirming if the simplification results in a determinate value.
These checks can help ensure that the proper methods are used and the correct result is obtained. Reinforcing these cautious approaches in calculus problem-solving can significantly enhance accuracy and understanding.

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