Can any help me with this. I need to solve this without L'Hopital's rule. $\displaystyle \lim$$\displaystyle ($$\displaystyle \ln$$\displaystyle ($$\displaystyle \cos$$\displaystyle ($$\displaystyle x$$\displaystyle )$$\displaystyle )$$\displaystyle /$$\displaystyle x^2$$\displaystyle ) when x->0 Thanks in advance 2. ## Re: Limit of ln(cosx)/x^2 Originally Posted by Bokas Can any help me with this. I need to solve this without L'Hopital's rule. \displaystyle \lim$$\displaystyle ($$\displaystyle \ln$$\displaystyle ($$\displaystyle \cos$$\displaystyle ($$\displaystyle x$$\displaystyle )$$\displaystyle )$$\displaystyle /$$\displaystyle x^2$$\displaystyle )$ when x->0

We start from the 'infinite product'...

$\displaystyle \cos x= \prod_{n=0}^{\infty} (1-\frac{ 4 x^{2}}{\pi^{2} (2n+1)^{2}})$ (1)

... and from (1) we obtain...

$\displaystyle \ln \cos x = \sum_{n=0}^{\infty} \ln \{1-\frac{ 4 x^{2}}{\pi^{2} (2n+1)^{2}}\}$ (2)

Now if we compute the coefficient of the $\displaystyle x^{2}$ term of the McLaurin expansion of (2) we obtain...

$\displaystyle a_{2}= -\sum_{n=0}^{\infty} \frac{4}{\pi^{2} (2n+1)^{2}}= -\frac{4}{\pi^{2}}\ \frac{\pi^{2}}{8} = -\frac{1}{2}$ (3)

... so that is...

$\displaystyle \lim_{x \rightarrow 0} \frac{\ln \cos x}{x^{2}}= -\frac{1}{2}$ (4)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Limit of ln(cosx)/x^2

Thank you very much. You saved me

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# lim(lncosx/ ln(1 x^2)

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