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.

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


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

  • 2
    $\begingroup$ What is your question? Your steps look good. $\endgroup$ – Gibbs Sep 23 '18 at 15:55
  • 2
    $\begingroup$ Your solution is fine. $\endgroup$ – Mark Sep 23 '18 at 15:56
  • $\begingroup$ @Gibbs My question is that does my logic and proof seem correct ? but you seem to answer this already, thanks ! $\endgroup$ – Tuki Sep 23 '18 at 16:00
  • $\begingroup$ @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. $\endgroup$ – Gibbs Sep 23 '18 at 16:04

Yes, your solution is correct.


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