You can use the tangent line approximation to create a linear function that gives a really close answer.

Let’s put ##f(x) = x^4,## we want ##f(1.999)## so use x= 1.999 and the nearby point of tangency a = 2. We’ll need ##f'(x)=4x^3## too.

The linear approximation we want (see my other answer) is

##f(x) ~~ f(a) + f'(a)(x-a)##

##f(1.999) ~~ f(2) + f'(2)(1.999-2)##

##~~ 2^4 + 4*2^3*(-0.001) = 16 – 0.032 = 15.968##

You can compare to the actual exact result of ##1.999^4 = 15.968023992001, ##so we came pretty close!

Bonus insight: The error depends on higher derivatives and can be predicted in advance! dansmath strikes again, approximately! /