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Question 16 of 40

Q.Express ∣x+1∣<0.5|x + 1| < 0.5 in neighbourhood and interval form.

(OR)
State multiplication and division working rule of limit.
Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2020Subjective· 2mImportance★★★★★
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∣x+1∣<0.5|x+1|<0.5 is the 0.50.5-neighbourhood of −1-1: N(−1,0.5)N(-1,0.5); as an interval, (−1.5, −0.5)(-1.5,\,-0.5). (OR: product/quotient rules of limits.)

Neighbourhood and interval form of ∣x+1∣<0.5|x+1| < 0.5.

Write ∣x+1∣=∣x−(−1)∣|x + 1| = |x - (-1)|. So the condition says the distance of xx from −1-1 is less than 0.50.5:

  • Neighbourhood form: it is the δ\delta-neighbourhood of the point a=−1a = -1 with radius δ=0.5\delta = 0.5:

    N(−1, 0.5)={ x:∣x−(−1)∣<0.5 }N(-1,\ 0.5) = \{\, x : |x - (-1)| < 0.5 \,\}

  • Interval form: remove the modulus:

    ∣x+1∣<0.5  ⟺  −0.5<x+1<0.5  ⟺  −1.5<x<−0.5|x+1| < 0.5 \iff -0.5 < x + 1 < 0.5 \iff -1.5 < x < -0.5

    ∴ x∈(−1.5, −0.5)\therefore\ x \in (-1.5,\ -0.5)

OR — Multiplication and division working rules of limits. If lim⁡x→af(x)\lim_{x\to a} f(x) and lim⁡x→ag(x)\lim_{x\to a} g(x) both exist, then:

  • Multiplication (product) rule: …

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