Limits

[math_fun]

The function f(x) has a jump discontinuity at x=0, where lim _{(x -> 0^-)} = 1, lim _{(x -> 0⁺)} = 2, and f(0) = 2. f(x) is continuous at all other real values of x. Find the limit lim _{(x -> 1)} (f(x-1)+f(1-x)).

I substituted x=1 into f(x-1)+f(1-x), and ended up with 2*f(0) = 2*2 = 4.

The solution solves for the left-side limit and the right-side limit at x=1 of f(x-1)+f(1-x)), which are both equal to 3. Therefore, the answer is 3.

Can anyone explain why my method was incorrect? Did the solution use one-sided limits because you are kind of solving for the limit at z=0 of 2*f(z) (though not exactly the same because the one-sided limits of f(x-1) and f(1-x) are different)? Thank you.

 

[Dr.Peterson]

You have to take into account the one-sided limits; you appear to have ignored the fact that they are different. You can't just substitute.

When x approaches 1 from the right, x-1 approaches 0 from the right, while 1-x approaches zero from the left. Therefore f(x-1) approaches 2, while f(1-x) approaches 1, and their sum is 2 + 1 = 3. Similarly, when x approaches 1 from the left, the two limits are reversed, and you get 1 + 2 = 3.

Don't try to take shortcuts. This has nothing to do with 2 f(z)!