# Hard limit

zoki85
Calculate:

$$lim_{z\to 0}(\frac{z}{sin z})^{\frac{1}{z^2}$$

$$z\in \mathbb{C}$$

I tried using Laurent series expansion but it gets too messy.
I hope LaTex turns out properly.

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

Overall, would it approach 1?

Homework Helper
How about using the natural logarithm function ?

lalbatros
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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zoki85
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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:

$$lim_{z\to 0}\frac{1}{z^2}*ln \frac{z}{sin z}$$

What result do you get?

lalbatros
zoki85,

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

michel

tehno
Calculate:

$$lim_{z\to 0}(\frac{z}{sin z})^{\frac{1}{z^2}$$

$$z\in \mathbb{C}$$

I tried using Laurent series expansion but it gets too messy.
I hope LaTex turns out properly.

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

Laurent series not necessary

zoki85
$$\lim_{z\to 0}\left(\frac{z}{sin\ z}\right)^{\frac{1}{Z^2}}=e^{1/6}$$

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

lalbatros what do you get?

tehno
Easily!

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

Becouse of :

$$\lim_{\phi\to 0}(1+\phi)^{1/\phi}=e$$

We have:

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

Limit:
$$\lim_{z\to 0}\frac{sin(z)-z}{z^3}=-1/6$$
is easily obtainable if you know to apply L'Hospital rule ,and derivations for elementar holomorphic functions.

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lalbatros
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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zoki85
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:
$$lim_{z\to 0}\left(\frac{sinz}{z}\right)^{\frac{sinz}{z-sinz}}$$

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

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Homework Helper
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.

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

shouldn't it be
$$\lim_{z\to 0}\left(\frac{z}{sin(z)}\right)^{1/z^2}=1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{1}{z^2}\ln\left(\frac{\sin z}{z}\right)}\right) = 1/\left(e^{\lim_{z\to 0}\frac{\ln \sin z - \ln z}{z^2}\right)$$

or am i missing something here?

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zoki85
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" :
$$lim_{x\to 0}(1+x)^{\frac{1}{x}$$
to the function expression.
By the way,it took me while to see that too!

murshid_islam
i still don't see exactly how tehno got:
$$1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{sin(z)-z}{z^3}}\right)$$

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tehno
i still don't see exactly how tehno got:
$$1/\left(\lim_{z\to 0}\frac{sin(z)}{z}\right)^{1/z^2}=1/\left(e^{\lim_{z\to 0}\frac{sin(z)-z}{z^3}}\right)$$

Look:

$$\frac{sin\ z}{z}=1+\frac{sin(z)-z}{z}$$

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

Do you see e now?

murshid_islam
Do you see e now?
yeah, i see it now. thanks.