What is the Correct Solution for this Complex Trigonometric Limit?

[Zipi Damn]

$\lim_{z \to 0} \frac{sin z}{z(z+i)}$

I applied L'Hopital and I got:

$\lim_{z \to 0} \frac{cos z}{2z+i}=\frac{1}{i}$

Wolphram Alpha's solution is -i. What am I doing wrong?

 

[CAF123]

Convince yourself that 1/i and -i are the same.

 

[Zipi Damn]

I think I will not pass the test.

Thanks.

 

[HallsofIvy]

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


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