How to prove limits by using the formal definition (epsilon-delta)?? I was told to prove the limit of $10-2x$ equals $16$ as $x\to-3$ using the formal definition (epsilon-delta).
This is what I tried so far, but I keep getting stuck. Can you help me solve this?
If $|x-(-3)|<\delta$ then $|(10-2x)-16|<\epsilon$.
Let $\delta = \frac{\epsilon}{2}$.
If $|x+3|< \frac{\epsilon}{2}$ then $-\frac{\epsilon}{2} <x+3 <\frac{\epsilon}{2}$.
So, $-\epsilon < 2x-6 <\epsilon$.
 A: Well, these types of direct problems that are an application of the definition are always solved in the same way. You have to reverse the sentence of the definition of a limit in p of a funcion f:
x=/=p, |x-p|<delta ----> |f(x)-L|<épsilon
You'll always try this:
|f(x)-L|<épsilon <--->...<--->....<---> |x-p|<something in terms of épsilon.
So you proved that the definition applies, because for any épsilon, you found a delta in terms of épsilon(in fact it applies for any delta<this delta you found)
So stop here and try to do it by yourself.
E = Épsilon
|10-2x-16|<E
<--->
|-6-2x|<E
<--->
2|3+x|<E
<--->
|x-(-3)|<E/2
So, for any E, take a delta<=E/2, and then this is true:
|x-(-3)|<delta -----> |f(x)-f(-3)| < E
(the definition, it is true because of what I did in the beginning)
So it is continuous in -3
A: Let $\epsilon > 0$. Define $\delta := \frac{\epsilon}{2}$.
So, if
$$
|x-(-3)| < \frac{\epsilon}{2}
$$
then
$$
|(10-2x)-16| = |2x+6| = 2|x+3| = 2|x-(-3)| < \epsilon
$$
Therefore,
$$
\lim_{x\to-3} (10-2x) = 16
$$
as required.
