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Compute the value of $\displaystyle \lim_{z \to 0} \left(\dfrac{1}{z^2} - \dfrac{1}{\sin^2 z}\right)$

We have used expansion, but found no leadings. Also, we try to factorize, hoping the wonders of trigonometry identity here, but seems no lead. Could you help us?

Shane Dizzy Sukardy
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  • turn it into a single fraction, and try to factor out a powers of z (sin(z) does have a factor of z from the power series) from both the numerator and denominator. – wfw Mar 14 '18 at 05:41
  • And then what is the next step? – Shane Dizzy Sukardy Mar 14 '18 at 05:47
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    Apply L'hopital's Rule twice after writing it as a single fraction and then look at the series expansion. – Kavi Rama Murthy Mar 14 '18 at 05:48
  • assuming you dont want to apply l'hopital's, put it as a single fraction and write out the first couple terms of the expansions of the numerator and denominator. $z^4$ can be factored from both, and then the limit is just the constant terms. – wfw Mar 14 '18 at 05:50

1 Answers1

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Hint:

$$\dfrac1{z^2}-\dfrac1{\sin^2z}=\dfrac{\sin z-z}{z^3}\left(\dfrac{\sin z}z+1\right)\left(\dfrac z{\sin z}\right)^2$$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion

lab bhattacharjee
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