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Exercise 9.6 · Q9

Q.If f(x)=x(−1)⌊1x⌋, x≤0f(x)=x(-1)^{\left\lfloor\frac1x\right\rfloor},\ x\le0, then the value of lim⁡x→0f(x)\displaystyle\lim_{x\to0}f(x) is equal to

(1) −1-1
(2) 00
(3) 22
(4) 44
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Step 1. For x<0x<0 near 00, ⌊1x⌋\left\lfloor\dfrac1x\right\rfloor is some integer nn, so (−1)n=±1(-1)^{n}=\pm1 always — its magnitude is exactly 11, however wildly it oscillates as x→0−x\to0^-.

Step 2. So ∣f(x)∣=∣x(−1)⌊1/x⌋∣=∣x∣⋅1=∣x∣|f(x)| = \left|x(-1)^{\lfloor1/x\rfloor}\right| = |x|\cdot1=|x|.

Step 3. This gives −∣x∣≤f(x)≤∣x∣-|x|\le f(x)\le|x| for all x≤0x\le0 near 00. …

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