MCQ
lf $\lim_\limits{\text{x}\rightarrow \text{a}}\text{f}(\text{x})=\text{L},$ then for eac $ \epsilon > 0$, there exists $ δ>0$ so that:
  • A
    $0<∣\text{x−a}∣<δ\Rightarrow ∣\text{f(x)}−L∣\geq \epsilon$
  • $0<∣\text{x−a}∣<δ\Rightarrow ∣\text{f(x)}−\text{L}∣<\epsilon$
  • C
    $ \text{a} < \text{x} < \text{a}+\delta\Rightarrow \text{f(x)} −\text{L}<\epsilon$
  • D
    $ a-\delta < x < a\Rightarrow |f(x)- L|<\epsilon$

Answer

Correct option: B.
$0<∣\text{x−a}∣<δ\Rightarrow ∣\text{f(x)}−\text{L}∣<\epsilon$
It is fundamental concept that,
for limit of a function $f(x)$ to exist at any point aa there exists a real number $δ>0,$
such that $0 < |x - a|< δ,$
for which $|f(x) - L| < \epsilon$ ,
where $\epsilon > 0$

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