you dont even need calculus to prove this.i teach this in my algebra classes.we have to first agree that "0.999..." is in fact a real number (this makes sense,since 0.999... is indeed a repeating decimal),and then all we need are the axioms of equality.

proof:let x=0.999...
=> 10(x)=10(0.999...) (by multiplication property of equality or MPE)
=> 10x=9.999...
=> 10x-0.999...=9.999...-0.999... (by addition property of equality or APE)
=> 10x-x=9.999...-0.999... (here we replaced the 0.999... on the left side by x,since they are equal)
=> 9x=9
=> x=1 (again,by MPE)
now,by transitivity of equality,since x is equal to both 0.999... and 1,then 0.999...=1.

QUOD ERAT DEMONSTRANDUM.

a few remarks:in this proof,please note that the addition and multiplication properties of equality only require that we add/multiply real numbers,regardless of whether these real numbers are terminating,infinitely repeating or non-terminating,non-repeating decimals.

and i have made this thread just a tiny bit boring.