Question #1a2d5
1 Answer
Aug 2, 2014
=1/2 Explanation,
=lim_(x->0)((e^x-1-x)/x^2) , plugging limit yields0/0 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