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Question Number 66379 by Sandy Suhendra last updated on 13/Aug/19
Commented by kaivan.ahmadi last updated on 13/Aug/19
=limx→π23cosx+5cos3xcot5x=hoplimx→.π2−3sinx−15sin3x−5(1+cot25x)=−3+15−5=−125
Commented by mathmax by abdo last updated on 14/Aug/19
letA(x)=(3cosx+5cos(3x))tan(5x)changementt=π2−xgivelimx→π2A(x)=limt→0(3sint+5cos(3π2−3t))tan(5π2−5t)=limt→0(3sint−5sin(3t))×1tan(5t)wehavesint∼t,sin(3t)∼3tandtan(5t)∼5t⇒3sint−5sin(3t)tan(5t)∼3t−15t5t=−125⇒limx→π2A(x)=−125
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