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Chapter 3: Problem 89

L'Hospital Rule Evaluate: \(\lim _{x \rightarrow 0}\left(\frac{e^{x}-1-x}{x^{2}}\right)\)

Short Answer

Expert verified
The limit as x approaches 0 of the function \(\left(\frac{e^{x}-1-x}{x^{2}}\right)\) is \(\frac{1}{2}\).

Step by step solution

01

Initial Limit Form

Rewrite the given limit expression to show that it is in the indeterminate form of type 0/0: \(\lim _{x \rightarrow 0}\left(\frac{e^{x}-1-x}{x^{2}}\right) = \frac{0}{0}\) which is an indeterminate form.
02

Apply L'Hopital's Rule

Since the limit is in the form 0/0, we can apply L'H么pital's Rule, which states that for functions in the indeterminate form 0/0, the limit of their ratio is equal to the limit of the ratio of their derivatives. We differentiate the numerator and the denominator to get: \(\lim _{x \rightarrow 0}\left(\frac{e^{x}-1-x}{x^{2}}\right) = \lim _{x \rightarrow 0}\left(\frac{e^{x}-1}{2x}\right)\)
03

Repeat L'Hopital's Rule

The resulting limit is again in the form 0/0, so we apply L'H么pital's Rule again and differentiate the numerator and the denominator: \(\lim _{x \rightarrow 0}\left(\frac{e^{x}-1}{2x}\right) = \lim _{x \rightarrow 0}\left(\frac{e^{x}}{2}\right)\)
04

Final Calculation

Now, we substitute x by 0, obtaining a determinant form that we can evaluate: \(\lim _{x \rightarrow 0}\left(\frac{e^{x}}{2}\right) = \frac{e^{0}}{2} = \frac{1}{2}\)

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Key Concepts

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

Indeterminate Forms
When dealing with limits, sometimes we encounter undefined expressions. These are known as indeterminate forms. One common type is \( \frac{0}{0} \). This occurs when both the numerator and denominator of a fraction approach zero as the variable approaches a certain value.
Indeterminate forms create challenges in calculation because they do not directly reveal the behavior of functions.
To resolve these, we often use techniques like algebraic manipulation or apply calculus-based rules.
  • Indeterminate forms indicate a need for further analysis.
  • In our example, the expression \(e^{x}-1-x\) and \(x^2\) both approach zero as \(x \to 0\).
  • Recognizing this form allows us to apply L'H么pital's Rule to find the limit.
Limit Evaluation
Evaluating limits is essential to understanding the behavior of functions as they approach a specific point. Often, algebraic simplification or substitution is sufficient, but not always.
When the expression is an indeterminate form like \( \frac{0}{0} \), L'H么pital's Rule becomes useful. This rule states that if the limit \( \frac{f(x)}{g(x)} \) is indeterminate, the limit is equal to \( \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \), the limit of their derivatives.
  • Use L'H么pital's Rule when direct substitution results in an indeterminate form.
  • Differentiate the numerator and denominator separately.
  • In our scenario, using L'H么pital's Rule twice lets us evaluate the limit effortlessly as \( x \to 0\).
Differentiation
Differentiation is the process of finding the derivative of a function. It's a fundamental concept in calculus that helps us understand how functions change.
In the context of L'H么pital's Rule, differentiation allows us to simplify limits when faced with indeterminate forms.
For example, in the limit \( \frac{e^{x}-1-x}{x^{2}} \), differentiating \( e^{x}-1 \) yields \( e^{x} \), and \( x^{2} \) yields \( 2x \). These give us a new expression to evaluate at the limit.
  • Differentiation gives the slope of the tangent line at any point on a curve.
  • It's useful for analyzing and manipulating complex functions like in our example.
  • Applying it twice here allows us to efficiently compute the limit of the given function.

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