- #1

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[tex]lim_{z\to 0}(\frac{z}{sin z})^{\frac{1}{z^2}[/tex]

[tex]z\in \mathbb{C}[/tex]

I tried using Laurent series expansion but it gets too messy.

I hope LaTex turns out properly.

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- Thread starter zoki85
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- #1

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[tex]lim_{z\to 0}(\frac{z}{sin z})^{\frac{1}{z^2}[/tex]

[tex]z\in \mathbb{C}[/tex]

I tried using Laurent series expansion but it gets too messy.

I hope LaTex turns out properly.

- #2

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Overall, would it approach 1?

- #3

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How about using the natural logarithm function ?

- #4

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Your idea:

Simply apply it to the expression within the bracket, see what happens and think to the definition of the exponential as a limit.

Remember: lim (1+1/n)^(x*n) = exp(x) for n->infinity

was very good.Laurent series expansion

Simply apply it to the expression within the bracket, see what happens and think to the definition of the exponential as a limit.

Remember: lim (1+1/n)^(x*n) = exp(x) for n->infinity

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- #5

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No.That's a typical error in thinking when you consider such limits.

Overall, would it approach 1?

Dextercioby,how to use natural logarithm function ?

I tried that too by taking the logarithm of the expression:

Can you calculate:

[tex]lim_{z\to 0}\frac{1}{z^2}*ln \frac{z}{sin z}[/tex]

What result do you get?

- #6

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zoki85,

develop the Ln(z/sin(z)) in series for small z, and you will get the result quickly,

michel

develop the Ln(z/sin(z)) in series for small z, and you will get the result quickly,

michel

- #7

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[tex]lim_{z\to 0}(\frac{z}{sin z})^{\frac{1}{z^2}[/tex]

[tex]z\in \mathbb{C}[/tex]

I tried using Laurent series expansion but it gets too messy.

I hope LaTex turns out properly.

[tex]\lim_{z\to 0}\left(\frac{z}{sin\ z}\right)^{\frac{1}{Z^2}}=e^{1/6}[/tex]

Laurent series not necessary

- #8

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Will you explain how to calculate the limit without Laurent please?[tex]\lim_{z\to 0}\left(\frac{z}{sin\ z}\right)^{\frac{1}{Z^2}}=e^{1/6}[/tex]

Laurent series not necessary

I used the Laurent,but I think I got different result than your .

lalbatros what do you get?

- #9

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Will you explain how to calculate the limit without Laurent please?

Becouse of :

[tex]\lim_{\phi\to 0}(1+\phi)^{1/\phi}=e[/tex]

We have:

[tex]\lim_{z\to 0}\left(\frac{z}{sin(z)}\right)^{1/z^2}=1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{sin(z)-z}{z^3}}\right)=e^{1/6}[/tex]

Limit:

[tex]\lim_{z\to 0}\frac{sin(z)-z}{z^3}=-1/6[/tex]

is easily obtainable if you know to apply L'Hospital rule ,and derivations for elementar holomorphic functions.

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- #10

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zoki85,

**one method**

z/sin(z) = z/(z-z³/6) = 1/(1-z²/6) = 1+z²/6 (up to 3th order)

therefore

(z/sin(z))^(1/z²) = (1+z²/6)^(1/z²) = (1+u)^(1/6/u) (by a change of variable with u->infinity)

and you get the result from tehno.

**second method**

Ln(z/sin(z)) = z²/6 (from the same series development)

therefore

ln((z/sin(z))^(1/z²)) = ln(z/sin(z)) * (1/z²) = 1/6

and the same result follows ...

z/sin(z) = z/(z-z³/6) = 1/(1-z²/6) = 1+z²/6 (up to 3th order)

therefore

(z/sin(z))^(1/z²) = (1+z²/6)^(1/z²) = (1+u)^(1/6/u) (by a change of variable with u->infinity)

and you get the result from tehno.

Ln(z/sin(z)) = z²/6 (from the same series development)

therefore

ln((z/sin(z))^(1/z²)) = ln(z/sin(z)) * (1/z²) = 1/6

and the same result follows ...

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- #11

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lalbatros and tehno:Thanks a lot ,3 solutions for 1 problem !

I had exam today (Complex Analysis) and guess what.One problem was to find:

[tex]lim_{z\to 0}\left(\frac{sinz}{z}\right)^{\frac{sinz}{z-sinz}}[/tex]

Your instructions were very helpful and I solved it (in two ways) with no problems.

Thanks once again.

I had exam today (Complex Analysis) and guess what.One problem was to find:

[tex]lim_{z\to 0}\left(\frac{sinz}{z}\right)^{\frac{sinz}{z-sinz}}[/tex]

Your instructions were very helpful and I solved it (in two ways) with no problems.

Thanks once again.

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- #12

Gib Z

Homework Helper

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- #13

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tehno said:[tex]\lim_{z\to 0}\left(\frac{z}{sin(z)}\right)^{1/z^2}=1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{sin(z)-z}{z^3}}\right)=e^{1/6}[/tex]

shouldn't it be

[tex]\lim_{z\to 0}\left(\frac{z}{sin(z)}\right)^{1/z^2}=1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{1}{z^2}\ln\left(\frac{\sin z}{z}\right)}\right) = 1/\left(e^{\lim_{z\to 0}\frac{\ln \sin z - \ln z}{z^2}\right)[/tex]

or am i missing something here?

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- #14

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It is just an application of the definition of "e" :

[tex]lim_{x\to 0}(1+x)^{\frac{1}{x}[/tex]

to the function expression.

By the way,it took me while to see that too!

- #15

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i still don't see exactly how tehno got:

[tex]1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{sin(z)-z}{z^3}}\right)[/tex]

[tex]1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{sin(z)-z}{z^3}}\right)[/tex]

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- #16

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i still don't see exactly how tehno got:

[tex]1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{sin(z)-z}{z^3}}\right)[/tex]

Look:

[tex]\frac{sin\ z}{z}=1+\frac{sin(z)-z}{z}[/tex]

[tex]\lim_{z\to 0}\left(\frac{sinz}{z}\right)^{1/z^2}=\lim_{z\to 0}\left(1+\left(\frac{sin(z)-z}{z}\right)\right)^{\left(\frac{z}{sin(z)-z}\right)\cdot \frac{sin(z)-z}{z^3}}[/tex]

Do you see

- #17

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yeah, i see it now. thanks.Do you seeenow?

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