How do you explain, using a complete sentence or sentences, why the calculation above is incorrect 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.

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ow do you explain, using a complete sentence or sentences, why the calculation above is incorrect

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

How do you explain, using a complete sentence or sentences, why the calculation above is incorrect

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.

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