i noticed it doesn't have x in them. i don't know what to do.
prove: let epsilon>0. let f(x) = pi + 3 for x∈ real number. Prove your conjecture. Whether f is
continuous for all r ∈ real number.
my proof;
|f(x)-f(r)|=|(pi+3)-(pi+3)|<epsilon
so 0<epsilon
thus f is continuous at 0. QED
To prove that a function is continuous for all real numbers, you need to show that the limit of the function exists and is equal to the value of the function at that point. In this case, we need to show that the function f(x) = pi + 3 is continuous for all real numbers.
To do this, we need to take the limit as x approaches r and show that it equals the value of the function at r.
In this case, the function f(x) is a constant function, as it is defined as pi + 3 for all x. Constant functions are always continuous for all real numbers.
To formally prove this, we can use the epsilon-delta definition of continuity. Let epsilon > 0 be given. We want to find a delta > 0 such that |f(x) - f(r)| < epsilon whenever |x - r| < delta.
For our function f(x) = pi + 3, we have |f(x) - f(r)| = |(pi + 3) - (pi + 3)| = 0.
No matter the value of epsilon, we can always choose a delta = 1, since for any |x - r| < 1, the difference |f(x) - f(r)| will always be 0.
Therefore, we have shown that for any given epsilon > 0, we can find a delta > 0 (in this case, delta = 1) such that |f(x) - f(r)| < epsilon whenever |x - r| < delta.
Hence, our function f(x) = pi + 3 is continuous for all real numbers. QED.