Q. 58 - Calculus

Question

Write delta-epsilon proofs for each of the limit statements  in Exercises $47–60.$

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

Step-by-Step Solution

Answer

Delta-epsilon proof is, 

Whenever $0<|x-2|<\delta$, we also have $\left|\frac{x^2-3x+2}{x-2}-1\right|<ϵ$.

Step 1. Given information

We are given, 

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

Step 2. Writing the delta-epsilon proofs

The strategy is to write delta-epsilon proofs for the given limit statement.

Consider that $\in >0$, choose $\delta =ϵ$.

For all x with $0<|x-2|<\delta$, we also have $\left|\frac{x^2-3x+2}{x-2}-1\right|<ϵ$.

$$\begin{gathered} \left|\frac{x^2-3x+2}{x-2}-1\right|=\left|\frac{x^2-3x+2-1(x-2)}{x-2}\right| \\ =\left|\frac{x^2-3x+2-x+2}{x-2}\right| \\ =\left|\frac{x^2-4x+4}{x-2}\right| \\ =\left|\frac{(x-2{)}^2}{x-2}\right| \\ =|x-2| \\ <\delta \\ =\in \end{gathered}$$

So, whenever $0<|x-2|<\delta$, we also have $\left|\frac{x^2-3x+2}{x-2}-1\right|<ϵ$.