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Complex trigonometric limit

  1. Jun 24, 2014 #1
    [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?
     
  2. jcsd
  3. Jun 24, 2014 #2

    CAF123

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    Gold Member

    Convince yourself that 1/i and -i are the same.
     
  4. Jun 24, 2014 #3
    I think I will not pass the test.

    Thanks.
     
  5. Jun 24, 2014 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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.
     
    Last edited: Jun 24, 2014
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