Q. 54 - Calculus

Question

Write delta-epsilon proofs for each of the limit statements $\underset{x\to c}{\lim }f\left(x\right)=L$ in Exercises $47-60$.

$\underset{x\to 3}{\lim }\left(x^2-6x+11\right)=2$.

Step-by-Step Solution

Answer

Delta-epsilon proofs for $\underset{x\to 3}{\lim }\left(x^2-6x+11\right)=2$ is, whenever $0<\left|x-3\right|<\delta$, we also have$\left|\left(x^2-6x+11\right)-2\right|<\epsilon$.

Step 1 . Given information

$\underset{x\to 3}{\lim }\left(x^2-6x+11\right)=2$.

Step 2 . Consider &#949; &#62; 0 , choose &#948; = &#949; .

For all $x$ with $0<\left|x-3\right|<\delta$, we also have $\left|\left(x^2-6x+11\right)-2\right|<\epsilon$.

$$\begin{gathered} \left|\left(x^2-6x+11\right)-2\right|=\left|x^2-6x+11-2\right| \\ =\left|x^2-6x+9\right| \\ =\left|{\left(x-3\right)}^2\right| \\ ={\left|x-3\right|}^2 \\ <{\left(\delta \right)}^2 \\ ={\left(\sqrt{\epsilon }\right)}^2 \\ =\epsilon \end{gathered}$$

Therefore, whenever $0<\left|x-3\right|<\delta$, we also have $\left|\left(x^2-6x+11\right)-2\right|<\epsilon$.