Q. 35 - Calculus

Question

For each limit statement , use algebra to find δ > 0 in terms of $\epsilon$ > 0 so that if 0 < |x − c| < δ, then | f(x) − L| < $\epsilon$.

$\underset{x\to 2}{\lim }(x^2-4x+6)=2$

Step-by-Step Solution

Answer

$\delta =\sqrt{\epsilon }$

Step1. Given information.

We have been given a limit statement as $\underset{x\to 2}{\lim }(x^2-4x+6)=2$.

We have to find $\delta \mathrm{in} \mathrm{terms} \mathrm{of} \epsilon$.

Step 2. Use algebra.

From the given limit statement, we can identify 

$$\begin{gathered} f\left(x\right)=x^2-4x+6 \\ c=2 \\ L=2 \\ \mathrm{For} \epsilon >0 \\ \left|\left(x^2-4x+6\right)-2\right|<\epsilon \\ \left|x^2-4x+6-2\right|<\epsilon \\ \left|x^2-4x+4\right|<\epsilon \\ \left|(x-2{)}^2\right|<\epsilon \\ |x-2{|}^2<\epsilon \\ |x-2|<\sqrt{\epsilon } \\ \mathrm{For} 0<|x-c|<\delta , \mathrm{we} \mathrm{get} |x-2|<\sqrt{\epsilon } \\ \mathrm{Therefore}, \delta =\sqrt{\epsilon } \end{gathered}$$