i=0;
while i<=20
  disp(i)
  i=i+1;
end

j=0;
while j<=2
  disp(j)
  j=j+0.1;
end

% This is technically the same as your code,
% so it should have suffered from exactly the same problem.
% But thanks to rearrangement of calculation,
% MATLAB will calculate the number of iterations is with a division,
% which gives ((20 - 0) / 0.1 + 1) == 201
% Because the result value of 201 is a "small" integer,
% the IEEE floating-point rounding logic will cause the result's
% representation to be exactly 201, and not some plus/minus epsilon.
%
% I have to emphasize this is not always the case; sometimes
% division can give not-exactly-integer results as well.
%
% This "almost, but not quite always" correct behavior led most
% beginning MATLAB users into thinking that MATLAB is less susceptible
% to the intricacies of finite-precision floating point arithmetics,
% but OP's code example shows that MATLAB users must exercise
% the same level of care as programmers of C.
%
for j = 0:0.1:20,
    disp(j);
end

% Integer index range is accurate up to `(2^53) - 1`
% beware of off-by-one.
idx = (0:200) * 0.1;
for j = idx,
    disp(j);
end

% If you only care about the start, stop,
% and the number of generated values.
% Beware of off-by-one.
idx = linspace(0, 20, 201);
for j = idx,
    disp(j);
end