Question #1a2d5

1 Answer
Gaurav
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#

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