# Question #1a2d5

##### 1 Answer

Aug 2, 2014

#=1/2#

Explanation,

#=lim_(x->0)((e^x-1-x)/x^2)# , plugging limit yields#0/0# formThis type of question, usually solve by using Taylor Series expansion.

Using Taylor Series expansion for

#e^x# , which is

#e^x=1+x+x^2/(2!)+x^3/(3!)+.........#

#=lim_(x->0)(((1+x+x^2/(2!)+x^3/(3!)+.........)-1-x)/x^2)#

#=lim_(x->0)((x^2/(2!)+x^3/(3!)+.........)/x^2)#

#=lim_(x->0)(1/(2!)+x/(3!)+.........)# Now, putting the limit, we get

#=1/(2!)=1/2#