Answer to Question #5022 in Real Analysis for Christine
As f is continuous function in c, then for each
epsilon1>0 there exists delta>0 such that for each x:
abs(c-x)<delta, then abs(f(c)-f(x))<epsilon1.
Let epsilon1 = epsilon /
2. Then there exists delta1 such that:
abs(f(c)-f(x))<epsilon/2 for each x
from (c-delta1, c+delta1). As f(c)>epsilon, then f(x) > epsilon / 2 >
Let a = c-delta1, b = c+delta1, (a,b) contains c. So, we get that for each
x from (a,b), f(x)>0.
The proof is the same for f(c)<0.
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