Prove $\lim_{x\rightarrow -3} 1-4x=13$

Question

Problem

Prove $$\lim_{x\rightarrow -3} 1-4x=13$$

Using $\delta, \epsilon$ definiton of limits.

Attempt to solve

I can use $\delta ,\epsilon$ definition of limit. If i can show

$$ |x-a| < \delta \implies |f(x)-L| < \epsilon$$

It implies limit exists according to $\delta, \epsilon$ definition of limits.

$$ |x-(-3)|< \delta \implies |1-4x-13|< \epsilon $$ $$ |1-4x-13|< \epsilon \iff |-4x-12| < \epsilon \iff |4x+12|< \epsilon $$ $$ |x-(-3)|< \delta \iff |x+3| < \delta $$ $$ \text{let } \delta = \epsilon/4$$ $$ |x+3| < \delta \implies |x+3|<\epsilon/4 \implies$$ $$ 4|x+3|<\epsilon \implies $$ $$ |4(x+3)|< \epsilon \implies $$ $$ |4x+12| < \epsilon $$

$$\tag*{$\square$}$$

I would like to have some feedback if my solution looks correct or not.

Answer 1

Yes, your solution is correct.