What is the Correct Solution for this Complex Trigonometric Limit?

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Homework Help Overview

The discussion revolves around evaluating the limit of a complex trigonometric expression as \( z \) approaches 0, specifically \(\lim_{z \to 0} \frac{\sin z}{z(z+i)}\). Participants are exploring the correct interpretation and calculation of this limit.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster applies L'Hopital's rule and arrives at a limit of \( \frac{1}{i} \), questioning the discrepancy with Wolfram Alpha's result of \(-i\). Other participants discuss the necessity of L'Hopital's rule and reference known limits from calculus, suggesting alternative approaches to the problem.

Discussion Status

Participants are engaged in clarifying the relationship between \( \frac{1}{i} \) and \(-i\), with some suggesting that the original poster's approach may not be necessary. There is an ongoing exploration of the limit's evaluation without reaching a definitive consensus.

Contextual Notes

Some participants express concern about their understanding and performance in the context of an upcoming test, indicating a potential pressure to grasp the material thoroughly.

Zipi Damn
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[itex]\lim_{z \to 0} \frac{sin z}{z(z+i)}[/itex]

I applied L'Hopital and I got:

[itex]\lim_{z \to 0} \frac{cos z}{2z+i}=\frac{1}{i}[/itex]

Wolphram Alpha's solution is -i. What am I doing wrong?
 
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Convince yourself that 1/i and -i are the same.
 
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I think I will not pass the test.

Thanks.
 
L'Hopital isn't really necessary here. One limit typically proved in Calculus I is [itex]\lim_{x\to 0} \frac{sin(x)}{x}= 1[/itex] for x real but it is easy to see that it is true for complex numbers also. So
[tex]\lim_{z\to 0}\frac{sin(z)}{z(z+1)}= \left(\lim_{z\to 0}\frac{sin(z)}{z}\right)\left(\lim_{z\to 0} \frac{1}{z+ i}\right)[/tex][tex]= (1)(1/(0+ i)= 1/i= -i[/tex].


Of course, i(-i)= -i^2= -(-1)= 1 so -i= 1/i.
 
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