lim-0 e^sinx - e^x/x
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Answer:
#(e^x-e^(sinx))/(x-sinx)= e^x((1-e^(sinx-x))/(x-sinx))#
now calling #h = sinx-x# we have
#(e^x-e^(sinx))/(x-sinx)=e^x((e^0-e^(0+h)))/(-h)=e^x(e^(0+h)-e^0)/h#
We know that #lim_(x->0)sinx/x=1 hArr lim_(x->0)(sinx-x)=0#
then
#lim_(x->0) rArr lim_(h->0)#
and then
#lim_(x->0)(e^x-e^(sinx))/(x-sinx)=(lim_(x->0)e^x)(lim_(h->0)(e^(0+h)-e^0)/h))=1*e'(0)=1#
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form me don't know this answer
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