Trigonometric limits

Posted December 13th, 2009 by alpha

$\mathop{\lim}\limits_{x \to 0}\frac{ln cosx}{x^2}$

Comments

On December 14th, 2009 Structure says:

$\mathop{\lim}\limits_{x \to 0}\frac{ln cosx}{x^2}$

$\lim_{x\to 0}\frac{-\frac{\sin x}{\cos x}}{2x}=-\frac{1}{2}$

I have used l'Hospital theorem.

On December 15th, 2009 alpha says:

Kind let me know more about it and its application.

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``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

On December 16th, 2009 Structure says:

$\lim_{x \to 0}(\cos x)^{\frac{1}{x^2}}=\lim_{x \to 0}(1-(1-\cos x))^{\frac{1}{1-\cos x}\lim_{x \to 0}\frac{1-\cos x}{x^2}}=e^{-\frac{1}{2}}$