Question #1a2d5

1 Answer
Aug 2, 2014

=1/2

Explanation,

=lim_(x->0)((e^x-1-x)/x^2), plugging limit yields 0/0 form

This 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