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Prove that the lim x→3   3x+4=13  using the formal definition of a limit.

 Feb 5, 2019

Best Answer 

 #2
avatar+4404 
+2

\(\forall \epsilon > 0 \text{ show }\exists \delta \ni |x-3| < \delta \Rightarrow |3x+4-13|< \epsilon\)

 

\(|3x+4-13| < \epsilon \Rightarrow\\ |3x-9| < \epsilon \Rightarrow \\ -\epsilon < 3x-9 < \epsilon \Rightarrow \\ 9-\epsilon < 3x < 9+\epsilon \Rightarrow \\ 3-\dfrac{\epsilon}{3} < x < 3+ \dfrac{\epsilon}{3}\\ |x-3|< \dfrac{\epsilon}{3}\)

 

\(\text{so choose }\delta = \dfrac{\epsilon}{3} \\ \text{and working everything backwards you end up with}\\ |3x+4-13| < \epsilon\\ \text{as desired}\)

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 Feb 5, 2019
 #1
avatar+79 
+1

Can't you just plug in the value 3 for x?

 

What do you mean by formal definition of a limit?

 Feb 5, 2019
 #3
avatar+271 
+1

I wish it was that easy

Ruublrr  Feb 6, 2019
 #2
avatar+4404 
+2
Best Answer

\(\forall \epsilon > 0 \text{ show }\exists \delta \ni |x-3| < \delta \Rightarrow |3x+4-13|< \epsilon\)

 

\(|3x+4-13| < \epsilon \Rightarrow\\ |3x-9| < \epsilon \Rightarrow \\ -\epsilon < 3x-9 < \epsilon \Rightarrow \\ 9-\epsilon < 3x < 9+\epsilon \Rightarrow \\ 3-\dfrac{\epsilon}{3} < x < 3+ \dfrac{\epsilon}{3}\\ |x-3|< \dfrac{\epsilon}{3}\)

 

\(\text{so choose }\delta = \dfrac{\epsilon}{3} \\ \text{and working everything backwards you end up with}\\ |3x+4-13| < \epsilon\\ \text{as desired}\)

Rom Feb 5, 2019
 #4
avatar+99230 
0

I'm sure you are correct Rom but I am with itsyaboi.

Talking about needing to go the long way around!

Melody  Feb 6, 2019
 #5
avatar+4404 
+1

well he did ask to use the formal definition of a limit  cheeky

Rom  Feb 6, 2019
 #6
avatar+99230 
0

Yes i understand that.    laugh

Sometimes I think mathematicians (not you) purposely design things to be complicated when it does not seem remotely necessary. 

Melody  Feb 6, 2019

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