Q. 67 - Calculus

Question

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

will have to bound $\delta$.

$\underset{x\to 3}{\lim }\left(x^2-2x-3\right)=0$

Step-by-Step Solution

Answer

The given limit $\underset{x\to 3}{\lim }\left(x^2-2x-3\right)=0$ is proved.

Step 1. Given information.

We are given,  

$\underset{x\to 3}{\lim }\left(x^2-2x-3\right)=0$

Step 2. Proving the limit

Given $\epsilon >0$, choose $\delta =\min \left(1,\frac{\epsilon }{5}\right)$.

Then if $0<|x-3|<\delta$, we have,

$$\begin{gathered} \left|\left(x^2-2x-3\right)-0\right|=\left|x^2-3x+x-3\right| \\ =\left|x\right(x-3)+1(x-3\left)\right| \\ =\left|\right(x-3\left)\right(x+1\left)\right| \\ =|x-3||x+1| \\ <\delta |x+1| \end{gathered}$$

Step 3. Proving the limit

Now from, $0<|x-3|<\delta$,

$$\begin{gathered} x-3<1 \\ x<1+3 \\ x<4 \end{gathered}$$

Therefore,

$$\begin{gathered} <\delta |x+1| \\ \le \delta \left(5\right) \\ \le \epsilon \end{gathered}$$

Hence Proved.