Question #57637

1 Answer
Andrea S.
Apr 12, 2017

#lim_(x->0^+) sinx lnx =0#

Explanation:

This limit is in the indeterminate form #0*oo#. To apply L'Hospital's rule we have to transform it in a way that it is in the form #0/0# or #oo/oo#:

#lim_(x->0^+) sinx lnx = lim_(x->0^+) lnx/(1/sinx)#

we can now apply the rule and have:

#lim_(x->0^+) lnx/(1/sinx) = lim_(x->0^+) (d/dx lnx)/(d/dx 1/sinx) = lim_(x->0^+) (1/x)/(-cosx/sin^2x) = lim_(x->0^+) -sinx /x tanx = -1*0 = 0#

graph{sinxlnx [-10, 10, -5, 5]}

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