Chapter 1: Problem 22
Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow 0} \frac{e^{x}-x-1}{x^{2}} $$
Short Answer
Expert verified
The limit evaluates to \( \frac{1}{2} \).
Step by step solution
01
Check the form of the limit
Evaluate the limit's expression \( \frac{e^{x} - x - 1}{x^2} \) as \( x \to 0 \). Both \( e^x - x - 1 \to 0 \) and \( x^2 \to 0 \) as \( x \to 0 \), which gives an indeterminate \( \frac{0}{0} \) form. This means we can apply L'Hôpital's rule.
02
Differentiate the numerator and the denominator
To apply L'Hôpital's rule, first differentiate the numerator \( e^{x} - x - 1 \) which gives \( e^{x} - 1 \), and the denominator \( x^2 \) which gives \( 2x \).
03
Apply L'Hôpital's Rule
Using L'Hôpital's Rule, evaluate the limit of the new expression:\[\lim_{x \to 0} \frac{e^{x} - 1}{2x}\]Substitute \( x = 0 \) again, we find it is still in the form \( \frac{0}{0} \).
04
Differentiate again
Differentiate the numerator \( e^{x} - 1 \) again to get \( e^{x} \) and differentiate \( 2x \) again to get \( 2 \).
05
Evaluate the new limit
With these derivatives, apply L'Hôpital's Rule again:\[\lim_{x \to 0} \frac{e^{x}}{2} = \frac{e^{0}}{2} = \frac{1}{2}\]
06
State the result
The original limit evaluates to \( \frac{1}{2} \) after applying L'Hôpital's Rule twice.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
When evaluating limits, we sometimes encounter what is called an indeterminate form. This can happen when direct substitution in a limit results in ambiguous expressions such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms are tricky because they do not provide clear information about the limit's behavior.
To handle these forms, we use special techniques. One common approach is L'Hôpital's Rule, which is particularly useful for dealing with the \( \frac{0}{0} \) indeterminate form. In our exercise, the limit \( \lim_{x \to 0} \frac{e^x - x - 1}{x^2} \) is in the form \( \frac{0}{0} \) when you try to evaluate it at \( x = 0 \).
This signals that we need to determine the limit using a more advanced method, such as rewriting, factoring, or differentiation, to get rid of the ambiguity and find the actual limit value. Understanding indeterminate forms is the first step in simplifying complex limit evaluations.
To handle these forms, we use special techniques. One common approach is L'Hôpital's Rule, which is particularly useful for dealing with the \( \frac{0}{0} \) indeterminate form. In our exercise, the limit \( \lim_{x \to 0} \frac{e^x - x - 1}{x^2} \) is in the form \( \frac{0}{0} \) when you try to evaluate it at \( x = 0 \).
This signals that we need to determine the limit using a more advanced method, such as rewriting, factoring, or differentiation, to get rid of the ambiguity and find the actual limit value. Understanding indeterminate forms is the first step in simplifying complex limit evaluations.
Differentiation
Differentiation is the process of finding the derivative of a function. It tells us how a function changes at any given point, essentially providing the function's rate of change. Differentiation is key in using L'Hôpital's Rule, as it involves differentiating both the numerator and the denominator of a fraction separately.
In the exercise, we differentiate the numerator \( e^x - x - 1 \) to get \( e^x - 1 \) and the denominator \( x^2 \) to get \( 2x \). This step transforms our indeterminate form into a more manageable one, allowing us to reevaluate the limit.
Since the first differentiation still results in an indeterminate form \( \frac{0}{0} \), we differentiate again. This time, the numerator \( e^x - 1 \) becomes \( e^x \) and the denominator \( 2x \) becomes \( 2 \). By using differentiation properly, we simplify the limit evaluation process, ultimately allowing us to resolve the ambiguity.
In the exercise, we differentiate the numerator \( e^x - x - 1 \) to get \( e^x - 1 \) and the denominator \( x^2 \) to get \( 2x \). This step transforms our indeterminate form into a more manageable one, allowing us to reevaluate the limit.
Since the first differentiation still results in an indeterminate form \( \frac{0}{0} \), we differentiate again. This time, the numerator \( e^x - 1 \) becomes \( e^x \) and the denominator \( 2x \) becomes \( 2 \). By using differentiation properly, we simplify the limit evaluation process, ultimately allowing us to resolve the ambiguity.
Limit Evaluation
Evaluating limits is crucial in mathematics, particularly for gauging the behavior of functions as variables approach specific points. It is a foundational concept in calculus, often used to analyze continuity and convergence. The original exercise aims to evaluate the limit \( \lim_{x \to 0} \frac{e^x - x - 1}{x^2} \) which, at first glance, seems complex due to its indeterminate form.
By employing L'Hôpital's Rule after differentiating both the numerator and the denominator, the indeterminate \( \frac{0}{0} \) form is resolved through further simplification. The evaluated new expression \( \lim_{x \to 0} \frac{e^{x}}{2} \) results in \( \frac{1}{2} \) when \( x = 0 \).
Thus, limit evaluation often requires turning an initial indeterminate form into a determinate result through techniques like L'Hôpital's Rule, highlighting an organized step-by-step approach to handling potential ambiguities.
By employing L'Hôpital's Rule after differentiating both the numerator and the denominator, the indeterminate \( \frac{0}{0} \) form is resolved through further simplification. The evaluated new expression \( \lim_{x \to 0} \frac{e^{x}}{2} \) results in \( \frac{1}{2} \) when \( x = 0 \).
Thus, limit evaluation often requires turning an initial indeterminate form into a determinate result through techniques like L'Hôpital's Rule, highlighting an organized step-by-step approach to handling potential ambiguities.