Q. 3 - Calculus

Question

Write down the formal delta-epsilon statement you would have to prove in order to prove the limit statement. 

$\underset{x\to -2}{lim}\frac{3}{x+1}=-3$

Step-by-Step Solution

Answer

For all epsilon positive there exists a delta positive if $0<|x+2|<\delta \mathrm{then} \left|\frac{3}{x+1}+3\right|<\epsilon$

Step 1. Given Information

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

Step 2. Explanation

From the given limit expression, we have,

$$\begin{gathered} f\left(x\right)=\frac{3}{x+1} \\ c=-2 \\ L=-3 \end{gathered}$$

Hence, delta-epsilon statement will be as follows,

For all epsilon positive there exists a delta positive if 

$$\begin{gathered} 0<|x+2|<\delta \mathrm{then} \\ \left|\frac{3}{x+1}-(-3)\right|<\epsilon \\ \left|\frac{3}{x+1}+3\right|<\epsilon \end{gathered}$$