Q. 53 - Calculus

Question

For each limit $\underset{x\to c}{lim}f\left(x\right)=L$in Exercises 43–54, use graphs and algebra to approximate the largest value of $\delta$ such that if $x\in (c-\delta ,c)\cup (c,c+\delta ) \mathrm{then} \mathrm{f}\left(\mathrm{x}\right)\in (\mathrm{L}-\epsilon ,\mathrm{L}+\epsilon )\mathit{ }$

$\underset{x\to -1}{lim}\frac{x^2-2x-3}{x+1}=-4,\epsilon =1$

Step-by-Step Solution

Answer

The required value of $\delta =1$

Step 1. Given Information

The given expression is $\underset{x\to -1}{lim}\frac{x^2-2x-3}{x+1}=-4,\epsilon =1$

Step 2. Explanation

From the given expression, we have, $c=-1, L=-4$

The limit expression can be written as a formal statement as below,

For all epsilon positive, there exists a delta positive such that if $x\in (-1-\delta ,-1)\cup (-1,-1+\delta ) \mathrm{Then}\mathit{ }\frac{{\mathrm{x}}^2-2\mathrm{x}-3}{\mathrm{x}+1}\in (-4-\mathrm{\epsilon },-4+\mathrm{\epsilon })$

Now, the largest value of delta is given by,

$$\begin{gathered} \frac{x^2-2x-3}{x+1}=L+\epsilon \\ \frac{x^2-2x-3}{x+1}=-4+1 \\ x-3=-3 \\ x=0 \end{gathered}$$

Thus,

$$\begin{gathered} \delta =0+1 \\ \delta =1 \end{gathered}$$