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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Question
Comput each of the following limits
![The image contains a mathematical limit expression, which is commonly studied in calculus. The expression is:
\[ \lim_{{x \to 0}} \frac{e^x - 1}{\sin x} \]
This represents the limit of the fraction \(\frac{e^x - 1}{\sin x}\) as \(x\) approaches 0. In this problem, both the numerator \(e^x - 1\) and the denominator \(\sin x\) approach 0 as \(x\) approaches 0, resulting in an indeterminate form \( \frac{0}{0} \). To solve this limit, one could use techniques such as L'Hôpital's Rule, which involves differentiating the numerator and the denominator separately, or leveraging known series expansions for \(e^x\) and \(\sin x\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff24651e9-0a43-4fcf-8170-c5e34dc6dcab%2F3d2be8d7-33be-42a8-8b1c-c209da25c3b7%2Fxk23frg_processed.png&w=3840&q=75)
Transcribed Image Text:The image contains a mathematical limit expression, which is commonly studied in calculus. The expression is:
\[ \lim_{{x \to 0}} \frac{e^x - 1}{\sin x} \]
This represents the limit of the fraction \(\frac{e^x - 1}{\sin x}\) as \(x\) approaches 0. In this problem, both the numerator \(e^x - 1\) and the denominator \(\sin x\) approach 0 as \(x\) approaches 0, resulting in an indeterminate form \( \frac{0}{0} \). To solve this limit, one could use techniques such as L'Hôpital's Rule, which involves differentiating the numerator and the denominator separately, or leveraging known series expansions for \(e^x\) and \(\sin x\).
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