Evaluate
$$\lim_{x \to 0} \log_{\ \cos(3x)}\cos(2ix)$$
where $i=\sqrt{-1}$
How to go about?
Use of L'Hospital rule and expansion is prohibited.
2026-04-05 16:58:54.1775408334
Complex Trigonometric Limit
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HINT:
As $\cos(ix)=\cosh(x)$
we have $$\lim_{x\to0}\dfrac{\ln(\cosh2x)}{\ln(\cos3x)}=\lim_{x\to0}\dfrac{\ln(1+\sinh^22x)}{\ln(1-\sin^23x)}=-\lim_{x\to0}\dfrac{\ln(1+\sinh^22x)}{\sinh^22x}\cdot\lim_{x\to0}\dfrac{-\sin^23x}{\ln(1-\sin^23x)}$$ $$\cdot4\lim_{x\to0}\left(\dfrac{\sinh2x}{2x}\right)^2\cdot\dfrac1{9\left(\dfrac{\sin3x}{3x}\right)^2}$$