Immediate. Immediate. Please help with limit problem?

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Answer 1 · 2018-03-31T13:52:51

#lim_(x->0)(e^x-e^-x)/x=2#

We want to solve

#lim_(x->0)(e^x-e^-x)/x#

This leads to an indeterminate form #0/0#,
therefore we can apply L'Hôpital's rule

#color(blue)(lim_(x->a)f(x)/g(x)=lim_(x->a)(f'(x))/(g'(x))#

Thus

#lim_(x->0)(e^x-e^-x)/x=lim_(x->0)(e^x+e^-x)/1=2#

Answer 2 · 2018-03-31T13:56:31

#2#

#e^x = sum_(k=0)^oo x^k/(k!)#
and
#e^-x = sum_(k=0)^oo (-x)^k/(k!)#

then #e^x-e^-x = 2sum_(k=0)^oo x^(2k+1)/((2k+1)!) =2x(1+f(x))#

then

#lim_(x->0)(e^x-e^-x)/x = lim_(x->0)2(x(1+f(x)))/x = lim_(x->0) 2(1+f(x))=2#

Answer 3 · 2018-04-01T03:29:35

Image....

my notebook..

Without L Hospital rule...