Q. 42 - Calculus

Question

For each limit statement in Exercises $41-44$, use algebra to find $\delta$ or $N$ in terms of $\epsilon$ or $M$, according to the appropriate formal limit definition.

$\underset{x\to \infty }{\lim }\frac{x-1}{x}=1$, find $N$ in terms of $\epsilon$.

Step-by-Step Solution

Answer

For $\underset{x\to \infty }{\lim }\frac{x-1}{x}=1$, $N=\frac{1}{\epsilon }$.

Step 1 . Given information

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

Step 2 . From the limit expression,

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

Now for $\epsilon >0$.

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

Therefore, for $x>N$, we get $\left|\left(\frac{x-1}{x}\right)-1\right|<\epsilon$.

Hence, $N=\frac{1}{\epsilon }$.