Question #d54ed

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Answer 1 · 2017-03-11T20:33:52

Our goal is to rewrite this so that we can use

$lim_(thetararr0)sintheta/theta = lim_(thetararr0)theta/sintheta =1$

In this case, it looks like we must have $theta = 2x$

Note that

$(3x^2+2x)/(sin2x) = x/(sin2x)(3x+2)$

$= (2x)/(sin2x) sin(3x+2)/2$

Now we can evaluate the limit as $xrarr0$

$lim_(xrarr0) (3x^2+2x)/(sin2x) = lim_(xrarr0) (2x)/(sin2x) sin(3x+2)/2$

$lim_(xrarr0) (2x)/(sin2x) lim_(xrarr0) sin(3x+2)/2$

$= (1)(2/2)=1$