We use the trig identity and then use L’Hopital’s rule to evaluate,
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Anonymous says:
I have a simple solution without using the L’hôpital rule for inf/inf.
log|senx| = log|sen2x| – ln|2cosx| , then divide by log|sen2x| and you have :
1-lim(ln|2cosx|/ln|sen2x| and the solving that easy limit you have 1
Eiji says:
My understanding is that the book has not proved to use the L’Hopital’s rule when the indeterminate form is ∞/∞. You need to show that the equation is lim x -> π 1/((log 2)/(log (sin x))+1+log(cos x)/log (sin x))
Anonymous says:
I agree, my solution is to set t = |sin(x)| and note that t->0+ when x->pi and then apply L’Hopital.
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![\[ \lim_{x \to \pi} \frac{\log |\sin x|}{\log | \sin (2x) |}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f335c9127155da94054dc706d4181364_l3.png "Rendered by QuickLaTeX.com")
and then use L’Hopital’s rule to evaluate,![\begin{align*} \lim_{x \to \pi} \frac{\log |\sin x|}{\log |\sin (2x)|} &= \lim_{x \to \pi} \frac{\log |\sin x|}{\log |2 \sin x \cos x|} \\[9pt] &= \lim_{x \to \pi} \frac{\log |\sin x|}{\log 2 + \log |\sin x| + \log |\cos x|} \\[9pt] &= \lim_{x \to \pi} \frac{ \frac{\cos x}{\sin x} }{ \frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}} &(\text{L'Hopital's}) \\[9pt] &= \lim_{x \to \pi} \frac{\cos x}{\cos x - \frac{\sin x}{\cos^2 x}} \\[9pt] &= 1. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-da4bf3dfd54c90fa760b502ba968750c_l3.png "Rendered by QuickLaTeX.com")
I have a simple solution without using the L’hôpital rule for inf/inf.
log|senx| = log|sen2x| – ln|2cosx| , then divide by log|sen2x| and you have :
1-lim(ln|2cosx|/ln|sen2x| and the solving that easy limit you have 1
My understanding is that the book has not proved to use the L’Hopital’s rule when the indeterminate form is ∞/∞. You need to show that the equation is lim x -> π 1/((log 2)/(log (sin x))+1+log(cos x)/log (sin x))
I agree, my solution is to set t = |sin(x)| and note that t->0+ when x->pi and then apply L’Hopital.