Question

lf $\underset{x\to a^{+}}{lim}f\left(x\right)=L$, then for each $ϵ>0$, there exists $\delta >0$ so that

• A. $0<|x-a|<\delta \Rightarrow |f\left(x\right)-L|\ge ϵ$
• B. $0<|x-a|<\delta \Rightarrow |f\left(x\right)-L|<ϵ$
• C. $a<x<a+\delta \Rightarrow f\left(x\right)-L<ϵ$
• D. $a-\delta <x<a\Rightarrow |f\left(x\right)-L|<ϵ$

Answer 1

Answer: (B)

$0<|x-a|<\delta \Rightarrow |f\left(x\right)-L|<ϵ$

It is fundamental concept that, for limit of a function $f\left(x\right)$ to exist at any point $a$ there exists a real number $\delta >0$, such that $0<|x-a|<\delta$, for which $|f\left(x\right)-L|<ϵ$, where $ϵ>0$