limx→0log(cosx)tan(x2)
That's the limit I need to solve, please notice that I can't use L'Hospital's Rule, now I'll show you what I've done so far:
limx→0log(cosx)tan(x2)=limx→0log(cosx)sin(x2)cos(x2)=cos(x2)log(cosx)sin(x2)
I was thinking about transforming the sinx into a cosx using this identity sin(x+π/2)=cosx, but it hasn't worked so far.
Thanks in advance.
Answer
Note that
logcosxtanx2=12logcos2xtanx2=12log(1−sin2x)tanx2==12log(1−sin2x)sinx2sinx2x2x2tanx2→12⋅−1⋅1⋅1=−12
since the following standard limits hold
sinxx→1tanxx→1log(1+x)x→1⟹−log(1−x)−x→−1
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