Sunday, 6 November 2016

Need help to solve this limit (Calculus) without L'Hospital's Rule



limx0log(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:



limx0log(cosx)tan(x2)=limx0log(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(1sin2x)tanx2==12log(1sin2x)sinx2sinx2x2x2tanx212111=12



since the following standard limits hold



sinxx1tanxx1log(1+x)x1log(1x)x1


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