Calculate Limit of (z/sin z)^(1/z^2) as z->0

  • Thread starter zoki85
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In summary, the limit as z approaches 0 of the expression (z/sin z)^(1/z^2) is equal to e^(1/6), which can be calculated using either the Laurent series expansion or the natural logarithm function. This can be obtained by rewriting the expression and applying the definition of the exponential function. Using this method, the limit can be easily calculated without having to use the Laurent series expansion.
  • #1
zoki85
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Calculate:

[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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  • #2
The bit inside the brackets approaches 1 and the 1/z^2 approaches infinity.

Overall, would it approach 1?
 
  • #3
How about using the natural logarithm function ?
 
  • #4
Your idea:
Laurent series expansion
was very good.

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
{

theperthvan said:
The bit inside the brackets approaches 1 and the 1/z^2 approaches infinity.

Overall, would it approach 1?
No.That's a typical error in thinking when you consider such limits.


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
zoki85,

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

michel
 
  • #7
zoki85 said:
Calculate:

[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 :smile:
 
  • #8
tehno said:
[tex]\lim_{z\to 0}\left(\frac{z}{sin\ z}\right)^{\frac{1}{Z^2}}=e^{1/6}[/tex]


Laurent series not necessary :smile:
Will you explain how to calculate the limit without Laurent please?
I used the Laurent,but I think I got different result than your .:mad:

lalbatros what do you get?
 
  • #9
Easily!

zoki85 said:
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
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 ...
 
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  • #11
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.
 
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  • #12
Nice work :) btw if you want the z>0 to be under the lim, rather than next to it, write \lim instead of just lim in your tex.
 
  • #13
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
I think you don't see there's no "ln" once "e" is introduced in calculations.
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
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]
 
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  • #16
murshid_islam said:
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 e now?
 
  • #17
tehno said:
Do you see e now?
yeah, i see it now. thanks.
 

1. What does it mean to calculate the limit of a function?

Calculating the limit of a function means finding the value that a function approaches as the input approaches a specific value or goes towards infinity.

2. Why is finding the limit of a function important?

Finding the limit of a function is important because it helps us understand the behavior of the function near a specific point. It also allows us to determine if the function is continuous or has any discontinuities.

3. What does "z->0" mean in the given function?

In the given function, "z->0" indicates that the input variable, z, is approaching the value of 0. This is known as the limit point, and we are trying to find the limit of the function at this point.

4. How do you calculate the limit of a complex function like (z/sin z)^(1/z^2)?

To calculate the limit of a complex function like (z/sin z)^(1/z^2), we can use various techniques such as algebraic manipulation, L'Hopital's rule, and Taylor series expansion. The specific method used will depend on the complexity of the function and the tools available.

5. What is the limit of (z/sin z)^(1/z^2) as z->0?

The limit of (z/sin z)^(1/z^2) as z->0 is equal to 1. This can be found using algebraic manipulation and applying the properties of limits. However, the limit does not exist if z is approaching 0 from the imaginary axis, as the function oscillates and does not approach a single value.

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