Question

$f\left(x\right)=2x+1,a=1,l=3$ and $ϵ=0.001$, then $\delta >0$ satisfying $0<|x-a|<\delta$ such that $|f\left(x\right)-l|<ϵ$, is

• A. $0.0005$
• B. $0.005$
• C. $0.001$
• D. $0.0001$

Answer 1

The correct option is A $0.0005$

We have to find $\delta >0$, which satisfies $0<|x-a|<\delta$ such that $|f\left(x\right)-l|<ϵ$
$\because |f\left(x\right)-l|<ϵ$
$\Rightarrow -ϵ<f\left(x\right)-l<ϵ$
$\Rightarrow l-ϵ<f\left(x\right)<l+ϵ$
$\Rightarrow 3-ϵ<2x+1<3+ϵ$
$\Rightarrow -ϵ<2x-2<ϵ$
$\Rightarrow -\frac{ϵ}{2}<x-1<\frac{ϵ}{2}$
$\Rightarrow |x-1|<\frac{ϵ}{2}$
$\Rightarrow |x-a|<\frac{ϵ}{2}$
$\delta =\frac{ϵ}{2}=\frac{0.001}{2}=0.0005$
Hence, option A.