Chapter D, Problem 15E

To determine

To prove:

  $\underset{x\to -3}{\lim }\left(1-4x\right)=13$ using $\delta ,\epsilon$ definition.

Explanation of Solution

Given information:

  $\underset{x\to -3}{\lim }\left(1-4x\right)=13$

Proof:

Definition of limit is,

  $\underset{x\to a}{\lim }f\left(x\right)=l$

So for each $\epsilon >0,\exists \delta >0$ such that

  $0<\left|x-a\right|<\delta \Rightarrow \left|f\left(x\right)-l\right|<\epsilon$

let $\epsilon >0$ be given,

So for finding out $\delta >0$ corresponding,

To such that,

  $\begin{array}{l}0<\left|x-\left(-3\right)\right|<\delta \Rightarrow \left|\left(1-4x\right)-13\right|<\epsilon \\ 0<\left|x+3\right|<\delta \Rightarrow \left|-4x-12\right|<\epsilon \\ 0<\left|x+3\right|<\delta \Rightarrow \left|-4\left(x+3\right)\right|<\epsilon \\ 0<\left|x+3\right|<\delta \to \left(1\right)\Rightarrow \left|x+3\right|<\frac{\epsilon }{4}\end{array}$

Take, $\delta =\frac{\epsilon }{4}$

Now,

  $\begin{array}{l}=4\left|x+3\right|<4\delta \left(by\left(1\right)\right)\\ =4\frac{\epsilon }{4}=\epsilon \\ \therefore \left|\left(1-4x\right)-13\right|<\epsilon \end{array}$

Thus verification is completed

  $\delta =\frac{\epsilon }{4}$ Satisfied condition of limit.

By the definition of limit $\underset{x\to -3}{\lim }\left(1-4x\right)=13$

For graph:

  $\begin{array}{l}y=1-4x\\ 4x+y=1\\ \frac{x}{\left(\frac{1}{4}\right)}+\frac{y}{\left(1\right)}=1\\ \left|x-\left(-3\right)\right|<\delta \Rightarrow -3-\delta <-3+\delta \\ \left|\left(1-4x\right)-13\right|<\epsilon \Rightarrow \left|y-13\right|<\epsilon \\ 13-\epsilon <y<13+\epsilon \end{array}$

So graph is as below:

  

Hence, it is proved.