Q. 70 - Calculus

Question

Prove each of the limit statements in Exercises 67–72. You

will have to bound $\delta$.

$\underset{x\to 1}{\lim }\left(x^2-6x+7\right)=2$

Step-by-Step Solution

Answer

The given limit $\underset{x\to 1}{\lim }\left(x^2-6x+7\right)=2$ is proved. 

Step 1. Given information.

We are given,   

$\underset{x\to 1}{\lim }\left(x^2-6x+7\right)=2$

Step 2. Proving the limit

Given $\epsilon >0$, choose $\delta =\min \left(1,\frac{\epsilon }{3}\right)$.

Then if $0<|x-1|<\delta$, we have,

$$\begin{gathered} \left|\left(x^2-6x+7\right)-2\right|=\left|x^2-6x+5\right| \\ =\left|x\right(x-5)-1(x-5\left)\right| \\ =\left|\right(x-5\left)\right(x-1\left)\right| \\ =|x-5||x-1| \\ <\delta |x-5| \end{gathered}$$

Step 3. Proving the limit

Now from, $0<|x-1|<\delta$,

$$\begin{gathered} x-1<1 \\ x<1+1 \\ x<2 \end{gathered}$$

Therefore, 

$$\begin{gathered} <\delta |2-5| \\ \le \delta \left(3\right) \\ \le \left(\frac{\epsilon }{3}\right)3 \\ \le \epsilon \end{gathered}$$

Hence Proved.