'Lim x-->0[e^x-e^sinx]/x-sinx'?

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Answer 1 · 2017-07-08T09:26:01

#1#

#(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#