Let $f:\mathbb {R}\to \mathbb {R}$ be an odd function such that $\displaystyle \lim _{x\to 0^+}{f(x)}=2$. Find $\displaystyle \lim _{x\to 0^-}{f(x)}$. (Write down X if the limit does not exist.)

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Consider the defining property of an odd function: f is odd if and only if f(-x) = -f(x) for all x.

Given that the right-hand limit as x approaches 0 from the positive side exists and equals 2, i.e., lim_{x→0^+} f(x) = 2, we can examine the behavior near 0 from the left using the oddness.

Take a sequence x_n → 0^- (x_n < 0). Let y_n = -x_n > 0. Then y_n → 0^+ and f(-y_n) = -f(y_n) by oddness. But f(-y_n) = f(x_n), so f(x_n) = -f(y_n).

Since lim_{y→0^+} f(y) = 2, the values f(y_n) approach 2 as n→∞. Therefore f(x_n) = -f(y_n) → -2.

Thus the left-hand limit exists and equals -2: lim_{x→0^-} f(x) = -2.

This reasoning relies on the given oddness of f and the existence of the right-hand limit; without oddness, the left-hand limit could behave differently. The fact that the problem asks to write X if the limit does not exist is not applicable here, because the combination of oddness with a finite right-hand limit ensures a finite left-hand limit of the opposite sign.

In summary, the left-hand limit must be the negation of the right-hand limit due to odd symmetry, so it equals -2. If one were to imagine a potential pitfall, assuming the left limit equals +2 would violate f(-x) = -f(x) and contradict the given right-hand limit under oddness.

_MATH1013_1ABCD_2025 Subsection 3.1 (closed on 27 Sep)

Let \(f:\mathbb {R}\to \mathbb {R}\) be an odd function such that \(\displaystyle \lim _{x\to 0^+}{f(x)}=2\). Find \(\displaystyle \lim _{x\to 0^-}{f(x)}\). (Write down X if the limit does not exist.)

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