Question

What is the value of `lim_(x->0)(2^x-1-x)/x^2` without using L'Hopital's rule and expansion of series?

Answer 1

See below.

Explanation:

Assuming the question is

`lim_(x->0)(2^x-1-x)/x` because

`lim_(x->0)(2^x-1-x)/x^2 = oo`

calling `f(x) = 2^x-x` we have

`lim_(x->0)(2^x-1-x)/x = lim_(x->0)(f(0+x)-f(0))/x = f'(0) = ln 2-1`