Q. 64 - Calculus

Question

For each of the limit statements in Exercises 61-66, write a $\delta -M,N-ϵ$, or $N-M$ proof, according to the type of limit statement.
$\underset{x\to -\infty }{\lim }\frac{2x-1}{x}=2$

Step-by-Step Solution

Answer

The proof of given limit $\underset{x\to -\infty }{\lim }\frac{2x-1}{x}=2$ is,  

$N=\frac{1}{\epsilon }$ .

Step 1. Given information

We are given, 

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

Step 2. Writing the delta-epsilon proofs

The strategy is to use algebra to find N in terms of $\epsilon$,

From the given limit, we get

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

Given $\epsilon <0$, choose $N=\frac{1}{\epsilon }$.

If $x<N$, we have,

$$\begin{gathered} \left|\frac{2x-1}{x}-2\right|=\left|\frac{2x-1-2x}{x}\right| \\ =\left|\frac{-1}{x}\right| \\ =\left|\frac{1}{x}\right| \\ <\frac{1}{N}=\epsilon \end{gathered}$$

Hence, $N=\frac{1}{\epsilon }$.