Given f:(a, b) →ℝ and x ∈(a, b), consider the statements: (i) limh →0 |f(x+h)-f(x)|=0 and (ii) l

step 1 So we have f of x plus h minus f of x minus h. Absolute value is more equal than f of x plus h m

Given f:(a, b) →ℝ and x ∈(a, b), consider the statements:...

Given $f:(a, b) \rightarrow \mathbb{R}$ and $x \in(a, b)$, consider the statements: (i) $\lim _{h \rightarrow 0}$ $|f(x+h)-f(x)|=0$ and (ii) $\lim _{h \rightarrow 0}|f(x+h)-f(x-h)|=0$. Show that (i) always implies (ii). Give an example where (ii) holds but not (i).

Key Concepts

Construction of Counterexamples via Pointwise Modification

The counterexample demonstrates that even if the symmetric difference tends to zero, the one?sided differences may fail to vanish at a specific point if the function has been redefined there. This typically involves modifying the value at the point (a “point discontinuity”) while leaving the surrounding behavior unchanged, so that the symmetric difference still cancels out the discontinuity, but the one?sided limit |f(x+h) – f(x)| does not tend to zero.

Limit of a Function at a Point

This concept involves understanding the behavior of a function f as its argument approaches a particular point. It provides a way to formalize what it means for the values of f(x+h) to become arbitrarily close to f(x) as h approaches zero. In the given problem, the limit is used to test whether f does not change “too much” near x, capturing the idea of local behavior of a function.

Continuity and One?Sided Continuity

Continuity at a point requires that the function’s value at that point equals the limit as the input approaches that point. In statement (i), the limit |f(x+h) – f(x)| = 0 implies continuity at x (from each side) in the sense that small perturbations from x yield values arbitrarily close to f(x). However, a function might fail to be continuous at x if one of the one?sided limits does not agree with f(x), even if symmetric differences behave well.

Symmetric Difference and the Triangle Inequality

The symmetric difference f(x+h) – f(x–h) measures the change in the function when comparing two points equidistant from x. By applying the triangle inequality, one obtains |f(x+h) – f(x–h)| ? |f(x+h) – f(x)| + |f(x) – f(x–h)|. Thus, if both one?sided differences tend to zero (i.e. if f is continuous in the respective one?sided sense), then the symmetric difference will also tend to zero. This reasoning explains why statement (i) implies statement (ii).

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