The function f(x) has a jump discontinuity at x=0, where lim

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

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

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.

_{(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.

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