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

## 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.

• What is your question? Your steps look good. – Gibbs Sep 23 '18 at 15:55
• Your solution is fine. – Mark Sep 23 '18 at 15:56
• @Gibbs My question is that does my logic and proof seem correct ? but you seem to answer this already, thanks ! – Tuki Sep 23 '18 at 16:00
• @Tuki it is ok. When you post something, make sure to write a question explicitly. Otherwise it may be closed because it is not clear what you are asking. – Gibbs Sep 23 '18 at 16:04