Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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How do you explain, using a complete sentence or sentences, why the calculation above is incorrect
Transcribed Image Text:Evaluate \(\lim_{x \to 0} \frac{\cos(2x) + 1}{e^x}\) correctly.
![The image illustrates the process of finding the limit of a mathematical expression using L'Hôpital's Rule.
The original limit is given as:
\[
\lim_{{x \to 0}} \frac{\cos(2x) + 1}{e^x}
\]
Using L'Hôpital's Rule (denoted by \( H \)), which is applicable when you have an indeterminate form like \(\frac{0}{0}\), we take the derivatives of the numerator and the denominator:
\[
= \lim_{{x \to 0}} \frac{-2\sin(2x)}{e^x}
\]
Evaluating this limit, we get:
\[
= \frac{0}{1} = 0
\]
Thus, the limit is 0. This process involves differentiating the numerator and denominator and then substituting the limit value.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0d5aff39-e45a-4c07-96a6-62051f0388e5%2F6c0e6c53-b8cc-49fa-a3b8-cdf72aeeb72d%2Fe4r82t9_processed.png&w=3840&q=75)
Transcribed Image Text:The image illustrates the process of finding the limit of a mathematical expression using L'Hôpital's Rule.
The original limit is given as:
\[
\lim_{{x \to 0}} \frac{\cos(2x) + 1}{e^x}
\]
Using L'Hôpital's Rule (denoted by \( H \)), which is applicable when you have an indeterminate form like \(\frac{0}{0}\), we take the derivatives of the numerator and the denominator:
\[
= \lim_{{x \to 0}} \frac{-2\sin(2x)}{e^x}
\]
Evaluating this limit, we get:
\[
= \frac{0}{1} = 0
\]
Thus, the limit is 0. This process involves differentiating the numerator and denominator and then substituting the limit value.
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