Question #91010

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Answer 1 · 2016-05-03T23:15:26

$lim_(x->0)(tan(x)-x)/(xsin(x))=0$

Direct substitution of $x=0$ shows that the initial limit results in a $0/0$ indeterminate form, and thus we can apply L'hopital's rule.

$lim_(x->0)(tan(x)-x)/(xsin(x)) = lim_(x->0)(d/dx(tan(x)-x))/(d/dxxsin(x))$

$=lim_(x->0)(sec^2(x)-1)/(xcos(x)+sin(x))$

As this again results in a $0/0$ indeterminate form, we may apply L'hopital's rule once again:

$=lim_(x->0)(d/dx(sec^2(x)-1))/(d/dx(xcos(x)+sin(x)))$

$=lim_(x->0)(2sec^2(x)tan(x))/(2cos(x)-xsin(x))$

$=(2sec^2(0)tan(0))/(2cos(0)-0sin(0))$

$=(2*1*0)/(2*1-0*0)$

$=0$