Question #553b2

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Answer 1 · 2016-12-15T14:22:12

$lim_(x->0)(1-cosx)/sin^2(3x) =1/18$

FIrst you can note that:

$(1-cosx)/sin^2(3x)= (1-cosx)/(3x)^2*((3x)/sin(3x))^2$

so that:

$lim_(x->0)(1-cosx)/sin^2(3x)=1/9*lim_(x->0)(1-cosx)/x^2*lim_(x->0)((3x)/sin(3x))^2$

But we know that:

$lim_(x->0)((3x)/sin(3x))^2 =1$

$lim_(x->0)(1-cosx)/x^2=1/2$

Finally:

$lim_(x->0)(1-cosx)/sin^2(3x) =1/18$