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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, $-\epsilon < 2x-6 <\epsilon$.

Arctic Char
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Ruby Padilla
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2 Answers2

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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

Ardov
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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.

Bob Zorro
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