Question

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

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

Answer 1

Answer: (B)

$0.0005$

$|f\left(x\right)-l|<0.001=ϵ$

$\Rightarrow |2x-3-1|<0.001$

$\Rightarrow -0.001<2x-4<0.001$
$\Rightarrow -0.0005<x-2<0.0005$
$\Rightarrow |x-2|<0.0005$
$\Rightarrow |x-a|<0.0005=\delta$

Hence, $\delta =0.0005>0$