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In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈. limx→2(2x+3) = 7

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प्रश्न

In the following example, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.

`lim_(x -> 2)(2x + 3)` = 7

योग
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उत्तर

Here, f(x) = 2x + 3, a = 2, l = 7

Let ∈ > 0 be given.

Consider |f(x) – l| < ∈

∴ |(2x + 3) – 7| < ∈

∴ |2x – 4| < ∈

∴ |2(x – 2)| < ∈

∴ 2|x – 2| < ∈

∴ `|x - 2| < ∈/2`

∴ if we take δ = `∈/2`, then

0 < |x – 2| < δ ⇒ |f(x) – 7| < ∈

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अध्याय 7: Limits - Exercise 7.1 [पृष्ठ १३९]

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