Immediate. Immediate. Please help with limit problem?

enter image source here

3 Answers
Mar 31, 2018

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

Explanation:

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

Mar 31, 2018

2

Explanation:

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

Apr 1, 2018

Image....

Explanation:

my notebook..my notebook..

Without L Hospital rule...